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Let’s say that $a$, $b$, and $c$ are integers such that $$(b^2+2)^2=(a^2+2c^2)(bc-a). \tag{$\star$}$$ By brute force search, I think I’ve discovered that $$(a,b,c)=(5d+1,3d+1,d+2), \qquad d=\dots,-2,-1,0,1,2,\dots$$ Certainly that condition satisfies ($\star$), since plugging in to either side gives $$81d^4+108d^3+90d^2+36d+9 = 81d^4+108d^3+90d^2+36d+9,$$ as desired.

QUESTION: How do I prove those are the only solutions to ($\star$)?

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EDIT: Observe that the existing “solution” is equivalent to saying that $$(a,b,c)=(5c-9,3c-5,c), \qquad c=\dots,-2,-1,0,1,2,\dots$$ are all solutions. That is to say, if $c$ is any integer satisfying ($\star$), then $(a,b)=(5c-9,3c-5)$ completes a solution of ($\star$). Does that consitute “necessary and sufficient conditions” on ($\star$), or do I need to show that there can be no other possible solutions?

Kieren MacMillan
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  • To answer your last question, in order to give necessary and sufficient conditions for $(\star)$ you must also prove those are the only solutions. I imagine it's quite common for an algebraic variety to admit parametrized solutions that don't cover all solutions. – Erick Wong May 30 '16 at 22:06
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    Here's a simple example that shows that there are more solutions than just the parametrized family you've found so far: $(-9, 5, 0)$. – Erick Wong May 30 '16 at 22:17
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    if you have a solution $(a,b,c)$, you get a new solution with $(a,-b,-c).$ All the solutions found so far either fit your pattern or do so if the involution is applied – Will Jagy Mar 11 '20 at 21:49

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