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 ![\[x^2+y^2 = 2014,\]](http://latex.artofproblemsolving.com/2/5/a/25a3179e00402d5b487e7101b600041bf5ffe906.png)
or prove that there are no solutions.
(b) Find all positive integers and that satisfy ![\[x^2 + y^2 = 3222014,\]](http://latex.artofproblemsolving.com/8/8/c/88cf74cb3ba0d63a5d843c9e2cfb0171dac5fa2f.png)
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 | ||
