Difference between revisions of "2006 Cyprus Seniors Provincial/2nd grade/Problem 4"

Latest revision as of 04:59, 12 November 2006

Find all integers pairs (x,y) that verify at the same time the inequalities $x^2\leq\frac{y^2+2x-1}{2}$ and $y^2\leq\frac{x^2-2y-1}{2}$ .

The inequality $x^2\leq\frac{y^2+2x-1}{2}$ becomes $2x^2\leq y^2+2x-1$ and the inequality $y^2\leq\frac{x^2-2y-1}{2}$ becomes $2y^2\leq x^2-2y-1$

By adding the inequalities we have $2x^2+2y^2 \leq y^2+2x-1+x^2-2y-1$

$x^2-2x+1+2y^2+2y+1 \leq 0$

$(x-1)^2+(y+1)^2 \leq 0$

But always $(x-1)^2+(y+1)^2 \geq 0$

So $(x-1)^2+(y+1)^2 = 0$

And the only integer pair is (1,-1)

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 == Solution ==
+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>
+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>
+<math>(x-1)^2+(y+1)^2 \leq 0</math>
+But always <math>(x-1)^2+(y+1)^2 \geq 0</math>
+So <math>(x-1)^2+(y+1)^2 = 0</math>
+And the only integer pair is (1,-1)