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- Feb 14, 2012
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Find all solutions to $(m^2+k)(m+k^2)=(m-k)^3$, where $m$ and $k$ are non-zero integers.
Find all solutions to $(m^2+k)(m+k^2)=(m-k)^3$, where $m$ and $k$ are non-zero integers.
Why is that?therefore $m(m-8) = (m-4)^2 - 16$ must also be a square. The only way that can happen is if $m-4 = \pm5$.
If $x^2-16 = y^2$ then $16 = x^2-y^2 = (x+y)(x-y)$. The only factorisation of $16$ giving positive integer values for $x$ and $y$ is $x+y=8$, $x-y=2$, so that $x=5$ and $y=3$.Why is that?therefore $m(m-8) = (m-4)^2 - 16$ must also be a square. The only way that can happen is if $m-4 = \pm5$.
Oops, I should have taken I like Serena's query more seriously. I was thinking that the only way you could have $x^2-16 = y^2$ was in the situation $5^2 - 4^2 = 3^2$. But in the solution to this problem, the situation $4^2 - 4^2 = 0^2$ is also possible. That leads to the solution $m=8$ and $k=-10$.Why is that?therefore $m(m-8) = (m-4)^2 - 16$ must also be a square. The only way that can happen is if $m-4 = \pm5$.