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 and that satisfy or prove that there are no solutions.
(b) Find all positive integers and that satisfy or prove that there are no
solutions.
Solution
(a) We see that we can rewrite as . Since and are perfect squares, their modulo can only be . Since none of those two combinations make , there are no solutions to such that .
(b) Similarly, we can rewrite as and therefore it also does not have integer solutions.
See Also
2014 UMO (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All UMO Problems and Solutions |