Difference between revisions of "2014 UMO Problems/Problem 2"

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== Solution ==
 
== Solution ==
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(a) We see that we can rewrite <math>x^2 + y^2 = 2014</math> as <math>x^2 + y^2 \equiv 6 \bmod{8}</math>. Since <math>x^2</math> and <math>y^2</math> are perfect squares, their modulo can only be <math>{0,1,4}</math>. Since none of those two combinations make <math>6</math>, there are no solutions to <math>x^2 + y^2 = 2014</math> such that <math>x,y \in \mathbb Z</math>.
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(b) Similarly, we can rewrite <math>x^2 + y^2 = 3222014</math> as <math>x^2 + y^2 \equiv 6 \bmod{8}</math> and therefore it also does not have integer solutions.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 11:36, 1 December 2014

Problem

(a) Find all positive integers $x$ and $y$ that satisfy \[x^2+y^2 = 2014,\] or prove that there are no solutions.


(b) Find all positive integers $x$ and $y$ that satisfy \[x^2 + y^2 = 3222014,\] or prove that there are no solutions.

Solution

(a) We see that we can rewrite $x^2 + y^2 = 2014$ as $x^2 + y^2 \equiv 6 \bmod{8}$. Since $x^2$ and $y^2$ are perfect squares, their modulo can only be ${0,1,4}$. Since none of those two combinations make $6$, there are no solutions to $x^2 + y^2 = 2014$ such that $x,y \in \mathbb Z$.

(b) Similarly, we can rewrite $x^2 + y^2 = 3222014$ as $x^2 + y^2 \equiv 6 \bmod{8}$ and therefore it also does not have integer solutions.

See Also

2014 UMO (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6
All UMO Problems and Solutions
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