How many integer solutions are there?

Algebra Level 2

How many integer solutions are there to the following expression?

x 2 + y 2 + z 2 = x 3 + y 3 + z 3 x^2+y^2+z^2=x^3+y^3+z^3

3 0 1 2 Infinitely many integer solutions

Hana Wehbi by Hana Wehbi , 51, USA

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2 solutions

Hana Wehbi
Sep 15, 2018

Set x = k ( 2 k 2 + 1 ) , y = 2 k 2 + 1 , z = − k ( 2 k 2 + 1 ) for k = 0 , 1 , 2 , … x= k(2k^2+1) , y= 2k^2+1, z= -k(2k^2+1) \text{ for } k=0,1,2,\dots

Since we only have to find infintely many solutions, not all of them, we can get rid of one variable and two cubes by taking z = − x z=-x . then we get the equation 2 x 2 + y 2 = y 3 2x^2+y^2=y^3 , which we can rewrite as x 2 = 1 2 ( y − 1 ) y 2 x^2=\frac{1}{2}(y-1)y^2 or x = ± y − 1 2 y . x=\pm \sqrt{\frac{y-1}{2}}y.

So if y − 1 2 \frac{y-1}{2} is a perfect square, x x is an integer. Thus, we take y − 1 2 = k 2 \frac{y-1}{2}=k^2 which gives y = 2 k 2 + 1 , x = k ( 2 k 2 + 1 ) y=2k^2+1, \ \ x= k(2k^2+1) Remark: \textit{Remark:} There certainly exist other solutions, e.g., x = 1 2 k ( k 2 − 1 ) + 1 , y = − 1 2 k ( k 2 − 1 ) + 1 , z = − k 2 + 1 for any k ∈ Z x=\frac{1}{2}k(k^2-1)+1,\ \ y=-\frac{1}{2}k(k^2-1)+1,\ \ z=-k^2+1 \text{ for any}\ k \in \mathbb{Z}

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The question is not difficult, because there are at least 8 easy solutions (when x, y, and z are all either 0 or 1), the question becomes much more difficult if you include the option '8' in the possible answers...

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