Maths Olympiad Prep

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Number theory Difficulty 4.9 AIME Prove it Soviet Union

Problem:
Find all integers xx, yy satisfying x2+x=y4+y3+y2+yx^{2} + x = y^{4} + y^{3} + y^{2} + y.

Solution

Solution:
The only solutions are x,y=1,1;0,1;0,0;6,2;5,2x, y = -1, 1; 0, -1; 0, 0; -6, 2; 5, 2.

(y2+y/21/2)(y2+y/2+1/2)=y4+y3+14y214<y4+y3+y2+y(y^{2} + y / 2 - 1 / 2)(y^{2} + y / 2 + 1 / 2) = y^{4} + y^{3} + \frac{1}{4}y^{2} - \frac{1}{4} < y^{4} + y^{3} + y^{2} + y except for 1y1/3-1 \leq y \leq -1/3.

Also (y2+y/2)(y2+y/2+1)=y4+y3+54y2+y/2(y^{2} + y / 2)(y^{2} + y / 2 + 1) = y^{4} + y^{3} + \frac{5}{4}y^{2} + y / 2 which is greater than y4+y3+y2+yy^{4} + y^{3} + y^{2} + y unless 0y20 \leq y \leq 2.

But no integers are greater than y2+y/21/2y^{2} + y / 2 - 1 / 2 and less than y2+y/2y^{2} + y / 2. So the only possible solutions have yy in the range 1-1 to 22. Checking these 4 cases, we find the solutions listed.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.