해밀턴 경로: 두 판 사이의 차이
내용 삭제됨 내용 추가됨
→예: 8×8=64 |
편집 요약 없음 |
||
44번째 줄:
| 날짜 = 2007
| isbn = 978-3-540-73544-1|언어=en}}</ref>
== 웹페이지를 사용할 수 없음 ==
다음 이유로 '''<nowiki>https://en.m.wikipedia.org/w/index.php?search=Sign+up+Vieta+Root+Jumping+Relevant+For...++Algebra+Vieta%27s+Formula+Algebra+Systems+of+Equations+Number+Theory+General+Diophantine+Equations+Sharky+Kesa%2C+Brandon+Monsen%2C+Satyabrata+Dash%2C+and++4+others++contributed+Vieta+jumping+is+a+nickname+for+a+particular+kind+of+descent+method+that+has+become+quite+popular+in+higher+level+math+Olympiad+number+theory+problems.+Like+other+instances+of+descent%2C+it+occurs+when+you+have+to+solve+a+Diophantine+equation+%28or+system+of+equations%2C+congruences+or+inequalities%29+whose+solutions+have+some+recursive+structure.+It+is+important+to+understand+that+Vieta+jumping+is+not+a+technique+that+you+can+choose+to+use+or+not+to+use%3A+most+Vieta+jumping+problems+can+only+be+solved+by+Vieta+jumping+%28else+would+require+a+tremendous+amount+of+work%29.++The+method+of+Vieta+jumping%2C+also+known+as+root+flipping%2C+can+be+very+useful+in+problems+involving+divisibility+of+positive+integers.+The+idea+is+to+assume+the+existence+of+a+solution+for+which+the+statement+in+question+is+wrong%2C+and+then+to+consider+the+given+relation+as+a+quadratic+equation+in+one+of+the+variables.+Using+Vieta%E2%80%99s+formula%2C+we+can+display+a+second+solution+to+this+equation.+The+next+step+is+to+show+that+the+new+solution+is+valid+and+smaller+than+the+previous+one.+Then+by+the+argument+of+infinite+descent+or+by+assuming+the+minimality+of+the+first+solution%2C+we+get+a+contradiction.+How+to+Recognize+Vieta+Jumping+is+just+a+particular+instance+of+a+more+general+method+of+descent%2C+so+don%27t+be+surprised+if+the+solutions+in+your+problem+have+a+more+complicated+recursive+structure.+The+tell-tale+sign+of+Vieta+jumping+is+the+existence+of+a+lemma+%28similar+in+spirit+to+the+one+below%29%2C+which+gives+a+relationship+between+different+possible+solutions.+Look+for+a+symmetric+equation+or+congruence+that+has+degree+two+in+each+of+the+two+variables.+%28Note+that+there+may+also+be+other+%22parameter%22+variables+in+the+problem.%29+Once+such+a+lemma+is+recognized+and+established%2C+you+should+look+for+some+inequalities+that+will+tell+you+in+what+direction+to+jump.++Usually%2C+the+lemma+will+say+something+along+the+lines+of+%22if++is+a+solution+for+some+constraint%2C+then++is+also+a+solution+for+the+same+constraint.%22+This+constraint+is+usually+given+by+fixing+one+or+more+of+the+variables.++In+the+case+of+a+quadratic%2C+we+can+consider+one+such+as+++which+is+a+quadratic+in+terms+of+.+We+know+the+properties+of+quadratics%2C+i.e.+there+are+always+two+solutions.+Hence%2C+if+we+fix++and+let+%2C+then+we+are+left+with+++And+we+can+use+Vieta%27s+formula+%28hence+the+name%29+to+get+a+relationship+between+the+two+solutions.+One+we+already+know+is+%2C+and+the+other+%28call+it+%29+gives+us++and+.++Upon+showing+this+relationship%2C+it+is+generally+helpful+to+look+for+ways+to+constrain+the+values+of+%2C+such+as+by+letting+a+relationship+between++and++dictate+another+relationship+between+and++and+seeing+where+that+goes.++The+end+goal+depends+on+the+problem%2C+but+will+usually+result+in+you+being+able+to+restrict+the+values+you+need+to+search+down+to+a+few+numbers%2C+or+in+other+cases+show+a+contradiction+if+conditions+%28what+the+problem+is+asking+to+prove%29+aren%27t+met.++The+Vieta+root+jumping+technique+allows+you+to+approach+problems+like+this%3A++What+is+the+largest+integer+%2C+such+that+there+exist+2+non-negative+integers++satisfying+++++Hint%3A++gives+us+%2C+so+the+answer+is+at+least++Submit+your+answer+1988+IMO+Example+Perhaps+the+earliest+Vieta+jumping+problem+was+Problem+6+at+the+1988+International+Mathematical+Olympiad.+At+the+time%2C+it+was+widely+recognized+as+both+beautiful+and+exceedingly+hard%2C+because+the+technique+of+Vieta+jumping+was+not+widely+used.++IMO%2788%2F6+Let++and++be+positive+integers+such+that++divides++Show+that++is+a+perfect+square.+Suppose++Then+for+a+fixed+the+set+of+solutions+has+a+very+interesting+structure%2C+described+by+the+following+lemma%3A++Lemma%3A+Suppose++is+a+positive+integer+solution+of+the+equation+Then++also+is+a+solution.++Proof+of+lemma%3A+Consider+the+quadratic+equation+We+get+this+equation+by+fixing++and+replacing++with+.++For+a+given+solution+%2C+we+know+that++is+a+root+of+the+above+equation.+By+Vieta+%28sum+of+roots%29%2C+we+know+that+the+other+root+is+%2C+which+is+an+integer.+Using+the+product+of+roots%2C+the+other+root+can+be+written+as+.+++If++let+the+other+root+be+.+We+have++If++then+which+contradicts++being+the+root+of+the+equation.+If++then++so++is+a+perfect+square+and+we+are+done.+Otherwise%2C+we+can+keep+going%2C+using++instead+of+.+We+get+a+decreasing+sequence+of+positive+integers+so+that+for+all++the+pair+satisfies++Clearly%2C+this+sequence+cannot+continue+indefinitely%2C+so+at+some+point+one+of+the+numbers+will+become++proving+that++is+a+perfect+square.+++The+infinite+sequence++as+above+is+what+distinguishes+Vieta+jumping+from+other+descent+methods.+Note+that+each+consecutive+pair+is+a+solution%2C+and++and++are+two+roots+of+a+quadratic+equation+that+involves+.+The+condition+must+be+symmetric+and+the+sequence+must+be+in+some+sense+decreasing%2C+so+the+sequence+must+terminate.+Note+also+that+Vieta+jumping+works+both+ways%3A+one+can+recover+all+solutions+from+the+terminal+ones+by+%22jumping+back.%22+Worked+Examples+Let++and++be+positive+integers+such+that+.+Show+that+.++We+see+that+%2C++and++are+solutions.+This+suggests+that+the+solutions+are+related+to+each+other+by+Vieta+root+jumping.++If+%2C+then+we+have+%2C+and+thus+.+Otherwise%2C+suppose+that++is+a+solution.+Let+.++Consider+the+following+quadratic+equation++Then%2C+we+know+that++would+be+a+solution+to+this+quadratic+equation%2C+and+the+other+solution+is+.+Since+%2C+it+is+an+integer.+Since+%2C+we+get+a+smaller+solution+.++Hence%2C+starting+with+any+solution+%2C+we+can+always+backtrack+along+it+until+.+This+gives+us+%2C+or+that+.+Hence%2C++or+2.++The+value+of++stays+the+same+throughout.+We+can+calculate+that+if+%2C+we+get+%2C+and+if++we+still+get+.+Hence%2C+we+must+always+have+%2C+or+that+.++%5BCrux+Mathematicorum%5D++Suppose+++and++are+positive+integers+such+that+Show+that++is+a+perfect+square.++Suppose++For+fixed++and++consider+all+positive+integer+pairs++such+that+Then+this+set+has+recursive+structure+given+by+the+following+lemma%3A++Lemma%3A++Suppose++satisfies+the+equation++and+.+Then++also+satisfies+the+equation+where++is+plugged+in+for++and++is+plugged+in+for+++Proof+of+lemma%3A++The+equation+has++as+one+of+its+roots.+Because+++cannot+be++and+by+Vieta%27s+formula+the+other+root+is++++Now+suppose++is+not+a+perfect+square.+Staring+from++with++we+will+apply+the+lemma.+Note+that+cannot+be+negative%2C+otherwise+we+get+a+contradiction%3A+Since++is+not+a+perfect+square%2C+cannot+be++so++Also%2C++We+can+continue+and+get+an+infinite+decreasing+sequence+which+is+impossible.+Therefore%2C++must+be+a+perfect+square.++Find+the+number+of+pairs+of+non-negative+integers++such+that+%2C+%2C+and+.++We+will+first+prove+the+following+lemma%3A++Lemma%3A++Suppose+the+pair++is+such+that+%2C+%2C+and+.++Then+the+pair+satisfies+the+conditions+%2C+%2C+and+.++Proof+of+lemma%3A++The+first+two+conditions+are+obvious.+To+prove+the+third+one%2C+note+that++and+%2C+and+thus+.+++Now+starting+from+any+pair+satisfying+the+condition+above%2C+we+apply+the+lemma+several+times%2C+producing+smaller+and+smaller+pairs%2C+until+we+get+to+a+pair++for+some+.+Note+that+the+above+construction+can+be+uniquely+reversed%3A+the+pair++is+obtained+from+a+pair++by+setting+.+So+we+can+generate+all+pairs+satisfying+the+conditions+of+the+problem+from+pairs++for+.+Specifically%2C++++++++++++++++++++++.++Overall%2C+we+get++pairs+with+%2C+plus++pairs+with+%2C+plus++other+pairs%2C+for+a+grand+total+of+pairs.++%5BIMO+2007%2F5%5D+Let++and++be+positive+integers+such+that+.+Show+that+.++It+is+hard+to+see+how+to+apply+Vieta%27s+formula+directly%2C+since+there+is+no+easy+way+to+%22jump%22+from+solution++to+.+The+question+suggests+that++would+be+an+important+quantity+to+consider.++Suppose+that++satisfies+the+conditions+with+.+Observe+that+Fix+.+We+want+to+consider+all+positive+integer+solutions++to+.+Let++be+the+solution+pair+which+minimizes+the+sum+of+both+values.+Since+the+equation+is+symmetric%2C+we+may+assume+that+.++Consider+the+quadratic+equation+It+has+2+roots%2C+namely++and+.+By+the+minimality+of+%2C+we+get+that+%2C+or+that+%2C+which+contradicts+.++Hence%2C+this+shows+that+no+solution+is+possible.++If++and++are+positive+integers+such+that++is+an+integer%2C+then+.++Lemma+1%3A++Suppose+a+certain+pair+of+integers+is+a+solution+to+the+equation+.+Then+I+say+the+pair++is+also+an+integer+solution.++Proof%3A++Consider+the+equation+.+By+assumption%2C++is+one+of+the+roots.+Therefore%2C+by+Vieta%27s+formulas%2C+the+other+root+is+.+By+substituting+the+formula+for+%2C+this+equals+.+We+already+know+that+the+numerator+is+divisible+by++%28since++is+an+integer%29%2C+and+it%27s+clearly+also+divisible+by+.+Since+%2C+the+numerator+is+also+divisible+by+.+Hence+the+new+root+is+an+integer.+Hence%2C+is+also+an+integer+solution+to+the+equation.+++Lemma+2%3A++If+%2C+then+.++Proof%3A++Since+%2C+it+follows+that+.+Therefore%2C+Hence+.+++Now%2C+let++be+the+solution+to++such+that+the+value+of++is+minimized+%28but+nonnegative%29.+Then%2C+by+Lemmas+1+and+2%2C+we+must+have+%2C+otherwise+it+is+possible+to+get+a+smaller+value+of+.++If+%2C+then+.+If+%2C+then+.+Then++and+therefore++respectively.+If+%2C+then+we+have+.+Thus+.+Hence%2C++is+a+factor+of+17.+Therefore+%2C+and+therefore+.+Therefore%2C+the+maximum+possible+value+of++in+any+of+these+cases+is+5.+++Ariel+Gershon++See+this+problem.+Additional+Problems+1%29+Find+all+pairs+of+integers++such+that++is+also+an+integer.++2%29+%5BVietnam+2002%5D+For+which+integers+%2C+does+there+exist+infinitely+many+integer+solutions+to+++3%29+%5BPutnam+1933%5D+Prove+that+for+every+real+number+%2C+the+equation+++has+a+solution+in+which++are+all+integers+greater+than+.++4%29+%5BSam+Vandervelde%5D+Let++be+positive+integers+such+that+.+Then%2C++can+be+written+as+the+sum+of+2+positive+squares.++Note%3A+This+generalizes+to+4+or+more+variables.+However%2C+Lagrange+4-square+theorem+tells+us+that+the+case+of+5+or+more+variables+doesn%27t+produce+a+meaningful+condition.++5%29+Determine+all+pairs+of+positive+integers++such+that++is+a+positive+integer.+Practice+math+and+science+questions+on+the+Brilliant+iOS+app.+Practice+math+and+science+questions+on+the+Brilliant+Android+app.+++About+Careers+Testimonials+Help+Terms+Privacy+%C2%A9+Brilliant+2017++Excel+in+math%2C+science%2C+and+engineering+Install+Sign+up+Vieta+Root+Jumping+Relevant+For...++Algebra+Vieta%27s+Formula+Algebra+Systems+of+Equations+Number+Theory+General+Diophantine+Equations+Sharky+Kesa%2C+Brandon+Monsen%2C+Satyabrata+Dash%2C+and++4+others++contributed+Vieta+jumping+is+a+nickname+for+a+particular+kind+of+descent+method+that+has+become+quite+popular+in+higher+level+math+Olympiad+number+theory+problems.+Like+other+instances+of+descent%2C+it+occurs+when+you+have+to+solve+a+Diophantine+equation+%28or+system+of+equations%2C+congruences+or+inequalities%29+whose+solutions+have+some+recursive+structure.+It+is+important+to+understand+that+Vieta+jumping+is+not+a+technique+that+you+can+choose+to+use+or+not+to+use%3A+most+Vieta+jumping+problems+can+only+be+solved+by+Vieta+jumping+%28else+would+require+a+tremendous+amount+of+work%29.++The+method+of+Vieta+jumping%2C+also+known+as+root+flipping%2C+can+be+very+useful+in+problems+involving+divisibility+of+positive+integers.+The+idea+is+to+assume+the+existence+of+a+solution+for+which+the+statement+in+question+is+wrong%2C+and+then+to+consider+the+given+relation+as+a+quadratic+equation+in+one+of+the+variables.+Using+Vieta%E2%80%99s+formula%2C+we+can+display+a+second+solution+to+this+equation.+The+next+step+is+to+show+that+the+new+solution+is+valid+and+smaller+than+the+previous+one.+Then+by+the+argument+of+infinite+descent+or+by+assuming+the+minimality+of+the+first+solution%2C+we+get+a+contradiction.+How+to+Recognize+Vieta+Jumping+is+just+a+particular+instance+of+a+more+general+method+of+descent%2C+so+don%27t+be+surprised+if+the+solutions+in+your+problem+have+a+more+complicated+recursive+structure.+The+tell-tale+sign+of+Vieta+jumping+is+the+existence+of+a+lemma+%28similar+in+spirit+to+the+one+below%29%2C+which+gives+a+relationship+between+different+possible+solutions.+Look+for+a+symmetric+equation+or+congruence+that+has+degree+two+in+each+of+the+two+variables.+%28Note+that+there+may+also+be+other+%22parameter%22+variables+in+the+problem.%29+Once+such+a+lemma+is+recognized+and+established%2C+you+should+look+for+some+inequalities+that+will+tell+you+in+what+direction+to+jump.++Usually%2C+the+lemma+will+say+something+along+the+lines+of+%22if++is+a+solution+for+some+constraint%2C+then++is+also+a+solution+for+the+same+constraint.%22+This+constraint+is+usually+given+by+fixing+one+or+more+of+the+variables.++In+the+case+of+a+quadratic%2C+we+can+consider+one+such+as+++which+is+a+quadratic+in+terms+of+.+We+know+the+properties+of+quadratics%2C+i.e.+there+are+always+two+solutions.+Hence%2C+if+we+fix++and+let+%2C+then+we+are+left+with+++And+we+can+use+Vieta%27s+formula+%28hence+the+name%29+to+get+a+relationship+between+the+two+solutions.+One+we+already+know+is+%2C+and+the+other+%28call+it+%29+gives+us++and+.++Upon+showing+this+relationship%2C+it+is+generally+helpful+to+look+for+ways+to+constrain+the+values+of+%2C+such+as+by+letting+a+relationship+between++and++dictate+another+relationship+between+and++and+seeing+where+that+goes.++The+end+goal+depends+on+the+problem%2C+but+will+usually+result+in+you+being+able+to+restrict+the+values+you+need+to+search+down+to+a+few+numbers%2C+or+in+other+cases+show+a+contradiction+if+conditions+%28what+the+problem+is+asking+to+prove%29+aren%27t+met.++The+Vieta+root+jumping+technique+allows+you+to+approach+problems+like+this%3A++What+is+the+largest+integer+%2C+such+that+there+exist+2+non-negative+integers++satisfying+++++Hint%3A++gives+us+%2C+so+the+answer+is+at+least++Submit+your+answer+1988+IMO+Example+Perhaps+the+earliest+Vieta+jumping+problem+was+Problem+6+at+the+1988+International+Mathematical+Olympiad.+At+the+time%2C+it+was+widely+recognized+as+both+beautiful+and+exceedingly+hard%2C+because+the+technique+of+Vieta+jumping+was+not+widely+used.++IMO%2788%2F6+Let++and++be+positive+integers+such+that++divides++Show+that++is+a+perfect+square.+Suppose++Then+for+a+fixed+the+set+of+solutions+has+a+very+interesting+structure%2C+described+by+the+following+lemma%3A++Lemma%3A+Suppose++is+a+positive+integer+solution+of+the+equation+Then++also+is+a+solution.++Proof+of+lemma%3A+Consider+the+quadratic+equation+We+get+this+equation+by+fixing++and+replacing++with+.++For+a+given+solution+%2C+we+know+that++is+a+root+of+the+above+equation.+By+Vieta+%28sum+of+roots%29%2C+we+know+that+the+other+root+is+%2C+which+is+an+integer.+Using+the+product+of+roots%2C+the+other+root+can+be+written+as+.+++If++let+the+other+root+be+.+We+have++If++then+which+contradicts++being+the+root+of+the+equation.+If++then++so++is+a+perfect+square+and+we+are+done.+Otherwise%2C+we+can+keep+going%2C+using++instead+of+.+We+get+a+decreasing+sequence+of+positive+integers+so+that+for+all++the+pair+satisfies++Clearly%2C+this+sequence+cannot+continue+indefinitely%2C+so+at+some+point+one+of+the+numbers+will+become++proving+that++is+a+perfect+square.+++The+infinite+sequence++as+above+is+what+distinguishes+Vieta+jumping+from+other+descent+methods.+Note+that+each+consecutive+pair+is+a+solution%2C+and++and++are+two+roots+of+a+quadratic+equation+that+involves+.+The+condition+must+be+symmetric+and+the+sequence+must+be+in+some+sense+decreasing%2C+so+the+sequence+must+terminate.+Note+also+that+Vieta+jumping+works+both+ways%3A+one+can+recover+all+solutions+from+the+terminal+ones+by+%22jumping+back.%22+Worked+Examples+Let++and++be+positive+integers+such+that+.+Show+that+.++We+see+that+%2C++and++are+solutions.+This+suggests+that+the+solutions+are+related+to+each+other+by+Vieta+root+jumping.++If+%2C+then+we+have+%2C+and+thus+.+Otherwise%2C+suppose+that++is+a+solution.+Let+.++Consider+the+following+quadratic+equation++Then%2C+we+know+that++would+be+a+solution+to+this+quadratic+equation%2C+and+the+other+solution+is+.+Since+%2C+it+is+an+integer.+Since+%2C+we+get+a+smaller+solution+.++Hence%2C+starting+with+any+solution+%2C+we+can+always+backtrack+along+it+until+.+This+gives+us+%2C+or+that+.+Hence%2C++or+2.++The+value+of++stays+the+same+throughout.+We+can+calculate+that+if+%2C+we+get+%2C+and+if++we+still+get+.+Hence%2C+we+must+always+have+%2C+or+that+.++%5BCrux+Mathematicorum%5D++Suppose+++and++are+positive+integers+such+that+Show+that++is+a+perfect+square.++Suppose++For+fixed++and++consider+all+positive+integer+pairs++such+that+Then+this+set+has+recursive+structure+given+by+the+following+lemma%3A++Lemma%3A++Suppose++satisfies+the+equation++and+.+Then++also+satisfies+the+equation+where++is+plugged+in+for++and++is+plugged+in+for+++Proof+of+lemma%3A++The+equation+has++as+one+of+its+roots.+Because+++cannot+be++and+by+Vieta%27s+formula+the+other+root+is++++Now+suppose++is+not+a+perfect+square.+Staring+from++with++we+will+apply+the+lemma.+Note+that+cannot+be+negative%2C+otherwise+we+get+a+contradiction%3A+Since++is+not+a+perfect+square%2C+cannot+be++so++Also%2C++We+can+continue+and+get+an+infinite+decreasing+sequence+which+is+impossible.+Therefore%2C++must+be+a+perfect+square.++Find+the+number+of+pairs+of+non-negative+integers++such+that+%2C+%2C+and+.++We+will+first+prove+the+following+lemma%3A++Lemma%3A++Suppose+the+pair++is+such+that+%2C+%2C+and+.++Then+the+pair+satisfies+the+conditions+%2C+%2C+and+.++Proof+of+lemma%3A++The+first+two+conditions+are+obvious.+To+prove+the+third+one%2C+note+that++and+%2C+and+thus+.+++Now+starting+from+any+pair+satisfying+the+condition+above%2C+we+apply+the+lemma+several+times%2C+producing+smaller+and+smaller+pairs%2C+until+we+get+to+a+pair++for+some+.+Note+that+the+above+construction+can+be+uniquely+reversed%3A+the+pair++is+obtained+from+a+pair++by+setting+.+So+we+can+generate+all+pairs+satisfying+the+conditions+of+the+problem+from+pairs++for+.+Specifically%2C++++++++++++++++++++++.++Overall%2C+we+get++pairs+with+%2C+plus++pairs+with+%2C+plus++other+pairs%2C+for+a+grand+total+of+pairs.++%5BIMO+2007%2F5%5D+Let++and++be+positive+integers+such+that+.+Show+that+.++It+is+hard+to+see+how+to+apply+Vieta%27s+formula+directly%2C+since+there+is+no+easy+way+to+%22jump%22+from+solution++to+.+The+question+suggests+that++would+be+an+important+quantity+to+consider.++Suppose+that++satisfies+the+conditions+with+.+Observe+that+Fix+.+We+want+to+consider+all+positive+integer+solutions++to+.+Let++be+the+solution+pair+which+minimizes+the+sum+of+both+values.+Since+the+equation+is+symmetric%2C+we+may+assume+that+.++Consider+the+quadratic+equation+It+has+2+roots%2C+namely++and+.+By+the+minimality+of+%2C+we+get+that+%2C+or+that+%2C+which+contradicts+.++Hence%2C+this+shows+that+no+solution+is+possible.++If++and++are+positive+integers+such+that++is+an+integer%2C+then+.++Lemma+1%3A++Suppose+a+certain+pair+of+integers+is+a+solution+to+the+equation+.+Then+I+say+the+pair++is+also+an+integer+solution.++Proof%3A++Consider+the+equation+.+By+assumption%2C++is+one+of+the+roots.+Therefore%2C+by+Vieta%27s+formulas%2C+the+other+root+is+.+By+substituting+the+formula+for+%2C+this+equals+.+We+already+know+that+the+numerator+is+divisible+by++%28since++is+an+integer%29%2C+and+it%27s+clearly+also+divisible+by+.+Since+%2C+the+numerator+is+also+divisible+by+.+Hence+the+new+root+is+an+integer.+Hence%2C+is+also+an+integer+solution+to+the+equation.+++Lemma+2%3A++If+%2C+then+.++Proof%3A++Since+%2C+it+follows+that+.+Therefore%2C+Hence+.+++Now%2C+let++be+the+solution+to++such+that+the+value+of++is+minimized+%28but+nonnegative%29.+Then%2C+by+Lemmas+1+and+2%2C+we+must+have+%2C+otherwise+it+is+possible+to+get+a+smaller+value+of+.++If+%2C+then+.+If+%2C+then+.+Then++and+therefore++respectively.+If+%2C+then+we+have+.+Thus+.+Hence%2C++is+a+factor+of+17.+Therefore+%2C+and+therefore+.+Therefore%2C+the+maximum+possible+value+of++in+any+of+these+cases+is+5.+++Ariel+Gershon++See+this+problem.+Additional+Problems+1%29+Find+all+pairs+of+integers++such+that++is+also+an+integer.++2%29+%5BVietnam+2002%5D+For+which+integers+%2C+does+there+exist+infinitely+many+integer+solutions+to+++3%29+%5BPutnam+1933%5D+Prove+that+for+every+real+number+%2C+the+equation+++has+a+solution+in+which++are+all+integers+greater+than+.++4%29+%5BSam+Vandervelde%5D+Let++be+positive+integers+such+that+.+Then%2C++can+be+written+as+the+sum+of+2+positive+squares.++Note%3A+This+generalizes+to+4+or+more+variables.+However%2C+Lagrange+4-square+theorem+tells+us+that+the+case+of+5+or+more+variables+doesn%27t+produce+a+meaningful+condition.++5%29+Determine+all+pairs+of+positive+integers++such+that++is+a+positive+integer.+Practice+math+and+science+questions+on+the+Brilliant+iOS+app.+Practice+math+and+science+questions+on+the+Brilliant+Android+app.+++About+Careers+Testimonials+Help+Terms+Privacy+%C2%A9+Brilliant+2017++Excel+in+math%2C+science%2C+and+engineering+InstallSign+up+Vieta+Root+Jumping+Relevant+For...++Algebra+Vieta%27s+Formula+Algebra+Systems+of+Equations+Number+Theory+General+Diophantine+Equations+Sharky+Kesa%2C+Brandon+Monsen%2C+Satyabrata+Dash%2C+and++4+others++contributed+Vieta+jumping+is+a+nickname+for+a+particular+kind+of+descent+method+that+has+become+quite+popular+in+higher+level+math+Olympiad+number+theory+problems.+Like+other+instances+of+descent%2C+it+occurs+when+you+have+to+solve+a+Diophantine+equation+%28or+system+of+equations%2C+congruences+or+inequalities%29+whose+solutions+have+some+recursive+structure.+It+is+important+to+understand+that+Vieta+jumping+is+not+a+technique+that+you+can+choose+to+use+or+not+to+use%3A+most+Vieta+jumping+problems+can+only+be+solved+by+Vieta+jumping+%28else+would+require+a+tremendous+amount+of+work%29.++The+method+of+Vieta+jumping%2C+also+known+as+root+flipping%2C+can+be+very+useful+in+problems+involving+divisibility+of+positive+integers.+The+idea+is+to+assume+the+existence+of+a+solution+for+which+the+statement+in+question+is+wrong%2C+and+then+to+consider+the+given+relation+as+a+quadratic+equation+in+one+of+the+variables.+Using+Vieta%E2%80%99s+formula%2C+we+can+display+a+second+solution+to+this+equation.+The+next+step+is+to+show+that+the+new+solution+is+valid+and+smaller+than+the+previous+one.+Then+by+the+argument+of+infinite+descent+or+by+assuming+the+minimality+of+the+first+solution%2C+we+get+a+contradiction.+How+to+Recognize+Vieta+Jumping+is+just+a+particular+instance+of+a+more+general+method+of+descent%2C+so+don%27t+be+surprised+if+the+solutions+in+your+problem+have+a+more+complicated+recursive+structure.+The+tell-tale+sign+of+Vieta+jumping+is+the+existence+of+a+lemma+%28similar+in+spirit+to+the+one+below%29%2C+which+gives+a+relationship+between+different+possible+solutions.+Look+for+a+symmetric+equation+or+congruence+that+has+degree+two+in+each+of+the+two+variables.+%28Note+that+there+may+also+be+other+%22parameter%22+variables+in+the+problem.%29+Once+such+a+lemma+is+recognized+and+established%2C+you+should+look+for+some+inequalities+that+will+tell+you+in+what+direction+to+jump.++Usually%2C+the+lemma+will+say+something+along+the+lines+of+%22if++is+a+solution+for+some+constraint%2C+then++is+also+a+solution+for+the+same+constraint.%22+This+constraint+is+usually+given+by+fixing+one+or+more+of+the+variables.++In+the+case+of+a+quadratic%2C+we+can+consider+one+such+as+++which+is+a+quadratic+in+terms+of+.+We+know+the+properties+of+quadratics%2C+i.e.+there+are+always+two+solutions.+Hence%2C+if+we+fix++and+let+%2C+then+we+are+left+with+++And+we+can+use+Vieta%27s+formula+%28hence+the+name%29+to+get+a+relationship+between+the+two+solutions.+One+we+already+know+is+%2C+and+the+other+%28call+it+%29+gives+us++and+.++Upon+showing+this+relationship%2C+it+is+generally+helpful+to+look+for+ways+to+constrain+the+values+of+%2C+such+as+by+letting+a+relationship+between++and++dictate+another+relationship+between+and++and+seeing+where+that+goes.++The+end+goal+depends+on+the+problem%2C+but+will+usually+result+in+you+being+able+to+restrict+the+values+you+need+to+search+down+to+a+few+numbers%2C+or+in+other+cases+show+a+contradiction+if+conditions+%28what+the+problem+is+asking+to+prove%29+aren%27t+met.++The+Vieta+root+jumping+technique+allows+you+to+approach+problems+like+this%3A++What+is+the+largest+integer+%2C+such+that+there+exist+2+non-negative+integers++satisfying+++++Hint%3A++gives+us+%2C+so+the+answer+is+at+least++Submit+your+answer+1988+IMO+Example+Perhaps+the+earliest+Vieta+jumping+problem+was+Problem+6+at+the+1988+International+Mathematical+Olympiad.+At+the+time%2C+it+was+widely+recognized+as+both+beautiful+and+exceedingly+hard%2C+because+the+technique+of+Vieta+jumping+was+not+widely+used.++IMO%2788%2F6+Let++and++be+positive+integers+such+that++divides++Show+that++is+a+perfect+square.+Suppose++Then+for+a+fixed+the+set+of+solutions+has+a+very+interesting+structure%2C+described+by+the+following+lemma%3A++Lemma%3A+Suppose++is+a+positive+integer+solution+of+the+equation+Then++also+is+a+solution.++Proof+of+lemma%3A+Consider+the+quadratic+equation+We+get+this+equation+by+fixing++and+replacing++with+.++For+a+given+solution+%2C+we+know+that++is+a+root+of+the+above+equation.+By+Vieta+%28sum+of+roots%29%2C+we+know+that+the+other+root+is+%2C+which+is+an+integer.+Using+the+product+of+roots%2C+the+other+root+can+be+written+as+.+++If++let+the+other+root+be+.+We+have++If++then+which+contradicts++being+the+root+of+the+equation.+If++then++so++is+a+perfect+square+and+we+are+done.+Otherwise%2C+we+can+keep+going%2C+using++instead+of+.+We+get+a+decreasing+sequence+of+positive+integers+so+that+for+all++the+pair+satisfies++Clearly%2C+this+sequence+cannot+continue+indefinitely%2C+so+at+some+point+one+of+the+numbers+will+become++proving+that++is+a+perfect+square.+++The+infinite+sequence++as+above+is+what+distinguishes+Vieta+jumping+from+other+descent+methods.+Note+that+each+consecutive+pair+is+a+solution%2C+and++and++are+two+roots+of+a+quadratic+equation+that+involves+.+The+condition+must+be+symmetric+and+the+sequence+must+be+in+some+sense+decreasing%2C+so+the+sequence+must+terminate.+Note+also+that+Vieta+jumping+works+both+ways%3A+one+can+recover+all+solutions+from+the+terminal+ones+by+%22jumping+back.%22+Worked+Examples+Let++and++be+positive+integers+such+that+.+Show+that+.++We+see+that+%2C++and++are+solutions.+This+suggests+that+the+solutions+are+related+to+each+other+by+Vieta+root+jumping.++If+%2C+then+we+have+%2C+and+thus+.+Otherwise%2C+suppose+that++is+a+solution.+Let+.++Consider+the+following+quadratic+equation++Then%2C+we+know+that++would+be+a+solution+to+this+quadratic+equation%2C+and+the+other+solution+is+.+Since+%2C+it+is+an+integer.+Since+%2C+we+get+a+smaller+solution+.++Hence%2C+starting+with+any+solution+%2C+we+can+always+backtrack+along+it+until+.+This+gives+us+%2C+or+that+.+Hence%2C++or+2.++The+value+of++stays+the+same+throughout.+We+can+calculate+that+if+%2C+we+get+%2C+and+if++we+still+get+.+Hence%2C+we+must+always+have+%2C+or+that+.++%5BCrux+Mathematicorum%5D++Suppose+++and++are+positive+integers+such+that+Show+that++is+a+perfect+square.++Suppose++For+fixed++and++consider+all+positive+integer+pairs++such+that+Then+this+set+has+recursive+structure+given+by+the+following+lemma%3A++Lemma%3A++Suppose++satisfies+the+equation++and+.+Then++also+satisfies+the+equation+where++is+plugged+in+for++and++is+plugged+in+for+++Proof+of+lemma%3A++The+equation+has++as+one+of+its+roots.+Because+++cannot+be++and+by+Vieta%27s+formula+the+other+root+is++++Now+suppose++is+not+a+perfect+square.+Staring+from++with++we+will+apply+the+lemma.+Note+that+cannot+be+negative%2C+otherwise+we+get+a+contradiction%3A+Since++is+not+a+perfect+square%2C+cannot+be++so++Also%2C++We+can+continue+and+get+an+infinite+decreasing+sequence+which+is+impossible.+Therefore%2C++must+be+a+perfect+square.++Find+the+number+of+pairs+of+non-negative+integers++such+that+%2C+%2C+and+.++We+will+first+prove+the+following+lemma%3A++Lemma%3A++Suppose+the+pair++is+such+that+%2C+%2C+and+.++Then+the+pair+satisfies+the+conditions+%2C+%2C+and+.++Proof+of+lemma%3A++The+first+two+conditions+are+obvious.+To+prove+the+third+one%2C+note+that++and+%2C+and+thus+.+++Now+starting+from+any+pair+satisfying+the+condition+above%2C+we+apply+the+lemma+several+times%2C+producing+smaller+and+smaller+pairs%2C+until+we+get+to+a+pair++for+some+.+Note+that+the+above+construction+can+be+uniquely+reversed%3A+the+pair++is+obtained+from+a+pair++by+setting+.+So+we+can+generate+all+pairs+satisfying+the+conditions+of+the+problem+from+pairs++for+.+Specifically%2C++++++++++++++++++++++.++Overall%2C+we+get++pairs+with+%2C+plus++pairs+with+%2C+plus++other+pairs%2C+for+a+grand+total+of+pairs.++%5BIMO+2007%2F5%5D+Let++and++be+positive+integers+such+that+.+Show+that+.++It+is+hard+to+see+how+to+apply+Vieta%27s+formula+directly%2C+since+there+is+no+easy+way+to+%22jump%22+from+solution++to+.+The+question+suggests+that++would+be+an+important+quantity+to+consider.++Suppose+that++satisfies+the+conditions+with+.+Observe+that+Fix+.+We+want+to+consider+all+positive+integer+solutions++to+.+Let++be+the+solution+pair+which+minimizes+the+sum+of+both+values.+Since+the+equation+is+symmetric%2C+we+may+assume+that+.++Consider+the+quadratic+equation+It+has+2+roots%2C+namely++and+.+By+the+minimality+of+%2C+we+get+that+%2C+or+that+%2C+which+contradicts+.++Hence%2C+this+shows+that+no+solution+is+possible.++If++and++are+positive+integers+such+that++is+an+integer%2C+then+.++Lemma+1%3A++Suppose+a+certain+pair+of+integers+is+a+solution+to+the+equation+.+Then+I+say+the+pair++is+also+an+integer+solution.++Proof%3A++Consider+the+equation+.+By+assumption%2C++is+one+of+the+roots.+Therefore%2C+by+Vieta%27s+formulas%2C+the+other+root+is+.+By+substituting+the+formula+for+%2C+this+equals+.+We+already+know+that+the+numerator+is+divisible+by++%28since++is+an+integer%29%2C+and+it%27s+clearly+also+divisible+by+.+Since+%2C+the+numerator+is+also+divisible+by+.+Hence+the+new+root+is+an+integer.+Hence%2C+is+also+an+integer+solution+to+the+equation.+++Lemma+2%3A++If+%2C+then+.++Proof%3A++Since+%2C+it+follows+that+.+Therefore%2C+Hence+.+++Now%2C+let++be+the+solution+to++such+that+the+value+of++is+minimized+%28but+nonnegative%29.+Then%2C+by+Lemmas+1+and+2%2C+we+must+have+%2C+otherwise+it+is+possible+to+get+a+smaller+value+of+.++If+%2C+then+.+If+%2C+then+.+Then++and+therefore++respectively.+If+%2C+then+we+have+.+Thus+.+Hence%2C++is+a+factor+of+17.+Therefore+%2C+and+therefore+.+Therefore%2C+the+maximum+possible+value+of++in+any+of+these+cases+is+5.+++Ariel+Gershon++See+this+problem.+Additional+Problems+1%29+Find+all+pairs+of+integers++such+that++is+also+an+integer.++2%29+%5BVietnam+2002%5D+For+which+integers+%2C+does+there+exist+infinitely+many+integer+solutions+to+++3%29+%5BPutnam+1933%5D+Prove+that+for+every+real+number+%2C+the+equation+++has+a+solution+in+which++are+all+integers+greater+than+.++4%29+%5BSam+Vandervelde%5D+Let++be+positive+integers+such+that+.+Then%2C++can+be+written+as+the+sum+of+2+positive+squares.++Note%3A+This+generalizes+to+4+or+more+variables.+However%2C+Lagrange+4-square+theorem+tells+us+that+the+case+of+5+or+more+variables+doesn%27t+produce+a+meaningful+condition.++5%29+Determine+all+pairs+of+positive+integers++such+that++is+a+positive+integer.+Practice+math+and+science+questions+on+the+Brilliant+iOS+app.+Practice+math+and+science+questions+on+the+Brilliant+Android+app.+++About+Careers+Testimonials+Help+Terms+Privacy+%C2%A9+Brilliant+2017++Excel+in+math%2C+science%2C+and+engineering+Install+appSign+up+Vieta+Root+Jumping+Relevant+For...++Algebra+Vieta%27s+Formula+Algebra+Systems+of+Equations+Number+Theory+General+Diophantine+Equations+Sharky+Kesa%2C+Brandon+Monsen%2C+Satyabrata+Dash%2C+and++4+others++contributed+Vieta+jumping+is+a+nickname+for+a+particular+kind+of+descent+method+that+has+become+quite+popular+in+higher+level+math+Olympiad+number+theory+problems.+Like+other+instances+of+descent%2C+it+occurs+when+you+have+to+solve+a+Diophantine+equation+%28or+system+of+equations%2C+congruences+or+inequalities%29+whose+solutions+have+some+recursive+structure.+It+is+important+to+understand+that+Vieta+jumping+is+not+a+technique+that+you+can+choose+to+use+or+not+to+use%3A+most+Vieta+jumping+problems+can+only+be+solved+by+Vieta+jumping+%28else+would+require+a+tremendous+amount+of+work%29.++The+method+of+Vieta+jumping%2C+also+known+as+root+flipping%2C+can+be+very+useful+in+problems+involving+divisibility+of+positive+integers.+The+idea+is+to+assume+the+existence+of+a+solution+for+which+the+statement+in+question+is+wrong%2C+and+then+to+consider+the+given+relation+as+a+quadratic+equation+in+one+of+the+variables.+Using+Vieta%E2%80%99s+formula%2C+we+can+display+a+second+solution+to+this+equation.+The+next+step+is+to+show+that+the+new+solution+is+valid+and+smaller+than+the+previous+one.+Then+by+the+argument+of+infinite+descent+or+by+assuming+the+minimality+of+the+first+solution%2C+we+get+a+contradiction.+How+to+Recognize+Vieta+Jumping+is+just+a+particular+instance+of+a+more+general+method+of+descent%2C+so+don%27t+be+surprised+if+the+solutions+in+your+problem+have+a+more+complicated+recursive+structure.+The+tell-tale+sign+of+Vieta+jumping+is+the+existence+of+a+lemma+%28similar+in+spirit+to+the+one+below%29%2C+which+gives+a+relationship+between+different+possible+solutions.+Look+for+a+symmetric+equation+or+congruence+that+has+degree+two+in+each+of+the+two+variables.+%28Note+that+there+may+also+be+other+%22parameter%22+variables+in+the+problem.%29+Once+such+a+lemma+is+recognized+and+established%2C+you+should+look+for+some+inequalities+that+will+tell+you+in+what+direction+to+jump.++Usually%2C+the+lemma+will+say+something+along+the+lines+of+%22if++is+a+solution+for+some+constraint%2C+then++is+also+a+solution+for+the+same+constraint.%22+This+constraint+is+usually+given+by+fixing+one+or+more+of+the+variables.++In+the+case+of+a+quadratic%2C+we+can+consider+one+such+as+++which+is+a+quadratic+in+terms+of+.+We+know+the+properties+of+quadratics%2C+i.e.+there+are+always+two+solutions.+Hence%2C+if+we+fix++and+let+%2C+then+we+are+left+with+++And+we+can+use+Vieta%27s+formula+%28hence+the+name%29+to+get+a+relationship+between+the+two+solutions.+One+we+already+know+is+%2C+and+the+other+%28call+it+%29+gives+us++and+.++Upon+showing+this+relationship%2C+it+is+generally+helpful+to+look+for+ways+to+constrain+the+values+of+%2C+such+as+by+letting+a+relationship+between++and++dictate+another+relationship+between+and++and+seeing+where+that+goes.++The+end+goal+depends+on+the+problem%2C+but+will+usually+result+in+you+being+able+to+restrict+the+values+you+need+to+search+down+to+a+few+numbers%2C+or+in+other+cases+show+a+contradiction+if+conditions+%28what+the+problem+is+asking+to+prove%29+aren%27t+met.++The+Vieta+root+jumping+technique+allows+you+to+approach+problems+like+this%3A++What+is+the+largest+integer+%2C+such+that+there+exist+2+non-negative+integers++satisfying+++++Hint%3A++gives+us+%2C+so+the+answer+is+at+least++Submit+your+answer+1988+IMO+Example+Perhaps+the+earliest+Vieta+jumping+problem+was+Problem+6+at+the+1988+International+Mathematical+Olympiad.+At+the+time%2C+it+was+widely+recognized+as+both+beautiful+and+exceedingly+hard%2C+because+the+technique+of+Vieta+jumping+was+not+widely+used.++IMO%2788%2F6+Let++and++be+positive+integers+such+that++divides++Show+that++is+a+perfect+square.+Suppose++Then+for+a+fixed+the+set+of+solutions+has+a+very+interesting+structure%2C+described+by+the+following+lemma%3A++Lemma%3A+Suppose++is+a+positive+integer+solution+of+the+equation+Then++also+is+a+solution.++Proof+of+lemma%3A+Consider+the+quadratic+equation+We+get+this+equation+by+fixing++and+replacing++with+.++For+a+given+solution+%2C+we+know+that++is+a+root+of+the+above+equation.+By+Vieta+%28sum+of+roots%29%2C+we+know+that+the+other+root+is+%2C+which+is+an+integer.+Using+the+product+of+roots%2C+the+other+root+can+be+written+as+.+++If++let+the+other+root+be+.+We+have++If++then+which+contradicts++being+the+root+of+the+equation.+If++then++so++is+a+perfect+square+and+we+are+done.+Otherwise%2C+we+can+keep+going%2C+using++instead+of+.+We+get+a+decreasing+sequence+of+positive+integers+so+that+for+all++the+pair+satisfies++Clearly%2C+this+sequence+cannot+continue+indefinitely%2C+so+at+some+point+one+of+the+numbers+will+become++proving+that++is+a+perfect+square.+++The+infinite+sequence++as+above+is+what+distinguishes+Vieta+jumping+from+other+descent+methods.+Note+that+each+consecutive+pair+is+a+solution%2C+and++and++are+two+roots+of+a+quadratic+equation+that+involves+.+The+condition+must+be+symmetric+and+the+sequence+must+be+in+some+sense+decreasing%2C+so+the+sequence+must+terminate.+Note+also+that+Vieta+jumping+works+both+ways%3A+one+can+recover+all+solutions+from+the+terminal+ones+by+%22jumping+back.%22+Worked+Examples+Let++and++be+positive+integers+such+that+.+Show+that+.++We+see+that+%2C++and++are+solutions.+This+suggests+that+the+solutions+are+related+to+each+other+by+Vieta+root+jumping.++If+%2C+then+we+have+%2C+and+thus+.+Otherwise%2C+suppose+that++is+a+solution.+Let+.++Consider+the+following+quadratic+equation++Then%2C+we+know+that++would+be+a+solution+to+this+quadratic+equation%2C+and+the+other+solution+is+.+Since+%2C+it+is+an+integer.+Since+%2C+we+get+a+smaller+solution+.++Hence%2C+starting+with+any+solution+%2C+we+can+always+backtrack+along+it+until+.+This+gives+us+%2C+or+that+.+Hence%2C++or+2.++The+value+of++stays+the+same+throughout.+We+can+calculate+that+if+%2C+we+get+%2C+and+if++we+still+get+.+Hence%2C+we+must+always+have+%2C+or+that+.++%5BCrux+Mathematicorum%5D++Suppose+++and++are+positive+integers+such+that+Show+that++is+a+perfect+square.++Suppose++For+fixed++and++consider+all+positive+integer+pairs++such+that+Then+this+set+has+recursive+structure+given+by+the+following+lemma%3A++Lemma%3A++Suppose++satisfies+the+equation++and+.+Then++also+satisfies+the+equation+where++is+plugged+in+for++and++is+plugged+in+for+++Proof+of+lemma%3A++The+equation+has++as+one+of+its+roots.+Because+++cannot+be++and+by+Vieta%27s+formula+the+other+root+is++++Now+suppose++is+not+a+perfect+square.+Staring+from++with++we+will+apply+the+lemma.+Note+that+cannot+be+negative%2C+otherwise+we+get+a+contradiction%3A+Since++is+not+a+perfect+square%2C+cannot+be++so++Also%2C++We+can+continue+and+get+an+infinite+decreasing+sequence+which+is+impossible.+Therefore%2C++must+be+a+perfect+square.++Find+the+number+of+pairs+of+non-negative+integers++such+that+%2C+%2C+and+.++We+will+first+prove+the+following+lemma%3A++Lemma%3A++Suppose+the+pair++is+such+that+%2C+%2C+and+.++Then+the+pair+satisfies+the+conditions+%2C+%2C+and+.++Proof+of+lemma%3A++The+first+two+conditions+are+obvious.+To+prove+the+third+one%2C+note+that++and+%2C+and+thus+.+++Now+starting+from+any+pair+satisfying+the+condition+above%2C+we+apply+the+lemma+several+times%2C+producing+smaller+and+smaller+pairs%2C+until+we+get+to+a+pair++for+some+.+Note+that+the+above+construction+can+be+uniquely+reversed%3A+the+pair++is+obtained+from+a+pair++by+setting+.+So+we+can+generate+all+pairs+satisfying+the+conditions+of+the+problem+from+pairs++for+.+Specifically%2C++++++++++++++++++++++.++Overall%2C+we+get++pairs+with+%2C+plus++pairs+with+%2C+plus++other+pairs%2C+for+a+grand+total+of+pairs.++%5BIMO+2007%2F5%5D+Let++and++be+positive+integers+such+that+.+Show+that+.++It+is+hard+to+see+how+to+apply+Vieta%27s+formula+directly%2C+since+there+is+no+easy+way+to+%22jump%22+from+solution++to+.+The+question+suggests+that++would+be+an+important+quantity+to+consider.++Suppose+that++satisfies+the+conditions+with+.+Observe+that+Fix+.+We+want+to+consider+all+positive+integer+solutions++to+.+Let++be+the+solution+pair+which+minimizes+the+sum+of+both+values.+Since+the+equation+is+symmetric%2C+we+may+assume+that+.++Consider+the+quadratic+equation+It+has+2+roots%2C+namely++and+.+By+the+minimality+of+%2C+we+get+that+%2C+or+that+%2C+which+contradicts+.++Hence%2C+this+shows+that+no+solution+is+possible.++If++and++are+positive+integers+such+that++is+an+integer%2C+then+.++Lemma+1%3A++Suppose+a+certain+pair+of+integers+is+a+solution+to+the+equation+.+Then+I+say+the+pair++is+also+an+integer+solution.++Proof%3A++Consider+the+equation+.+By+assumption%2C++is+one+of+the+roots.+Therefore%2C+by+Vieta%27s+formulas%2C+the+other+root+is+.+By+substituting+the+formula+for+%2C+this+equals+.+We+already+know+that+the+numerator+is+divisible+by++%28since++is+an+integer%29%2C+and+it%27s+clearly+also+divisible+by+.+Since+%2C+the+numerator+is+also+divisible+by+.+Hence+the+new+root+is+an+integer.+Hence%2C+is+also+an+integer+solution+to+the+equation.+++Lemma+2%3A++If+%2C+then+.++Proof%3A++Since+%2C+it+follows+that+.+Therefore%2C+Hence+.+++Now%2C+let++be+the+solution+to++such+that+the+value+of++is+minimized+%28but+nonnegative%29.+Then%2C+by+Lemmas+1+and+2%2C+we+must+have+%2C+otherwise+it+is+possible+to+get+a+smaller+value+of+.++If+%2C+then+.+If+%2C+then+.+Then++and+therefore++respectively.+If+%2C+then+we+have+.+Thus+.+Hence%2C++is+a+factor+of+17.+Therefore+%2C+and+therefore+.+Therefore%2C+the+maximum+possible+value+of++in+any+of+these+cases+is+5.+++Ariel+Gershon++See+this+problem.+Additional+Problems+1%29+Find+all+pairs+of+integers++such+that++is+also+an+integer.++2%29+%5BVietnam+2002%5D+For+which+integers+%2C+does+there+exist+infinitely+many+integer+solutions+to+++3%29+%5BPutnam+1933%5D+Prove+that+for+every+real+number+%2C+the+equation+++has+a+solution+in+which++are+all+integers+greater+than+.++4%29+%5BSam+Vandervelde%5D+Let++be+positive+integers+such+that+.+Then%2C++can+be+written+as+the+sum+of+2+positive+squares.++Note%3A+This+generalizes+to+4+or+more+variables.+However%2C+Lagrange+4-square+theorem+tells+us+that+the+case+of+5+or+more+variables+doesn%27t+produce+a+meaningful+condition.++5%29+Determine+all+pairs+of+positive+integers++such+that++is+a+positive+integer.+Practice+math+and+science+questions+on+the+Brilliant+iOS+app.+Practice+math+and+science+questions+on+the+Brilliant+Android+app.+++About+Careers+Testimonials+Help+Terms+Privacy+%C2%A9+Brilliant+2017++Excel+in+math%2C+science%2C+and+engineering+Install+app+app</nowiki>'''의 웹페이지를 로드할 수 없습니다.
net::ERR_CONNECTION_RESET
== 예 ==
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