Difference between revisions of "2006 Cyprus Seniors Provincial/2nd grade/Problem 4"
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+ | The inequality <math>x^2\leq\frac{y^2+2x-1}{2}</math> becomes <math>2x^2\leq y^2+2x-1</math> and the inequality <math>y^2\leq\frac{x^2-2y-1}{2}</math> becomes <math>2y^2\leq x^2-2y-1</math> | ||
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+ | By adding the inequalities we have <math>2x^2+2y^2 \leq y^2+2x-1+x^2-2y-1</math> | ||
+ | |||
+ | <math>x^2-2x+1+2y^2+2y+1 \leq 0</math> | ||
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+ | <math>(x-1)^2+(y+1)^2 \leq 0</math> | ||
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+ | But always <math>(x-1)^2+(y+1)^2 \geq 0</math> | ||
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+ | So <math>(x-1)^2+(y+1)^2 = 0</math> | ||
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+ | And the only integer pair is (1,-1) | ||
Latest revision as of 04:59, 12 November 2006
Problem
Find all integers pairs (x,y) that verify at the same time the inequalities and .
Solution
The inequality becomes and the inequality becomes
By adding the inequalities we have
But always
So
And the only integer pair is (1,-1)