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I'm trying to prove the following result.

Conjecture. If $z$ is an integer, and $z^4-6z^2+4z-3$ is a square, then $z=3$.

A quick check modulo $9$ shows that $z=9w+3$ for some integer $w$. So for some integer $y$ we must have \begin{align} y^2 &= z^4-6z^2+4z-3 \\ &= (9w+3)^4 - 6(9w+3)^2 + 4(9w+3) - 3\\ &= 9(729w^4 + 972w^3 + 432w^2 + 76w + 4). \end{align} So $y = 3u$ for an integer $u$, and hence $$ u^2 = 729w^4 + 972w^3 + 432w^2 + 76w + 4. $$

From here, I can rearrange and factor as $$ (u-2)(u+2) = w(729w^3 + 972w^2 + 432w + 76) $$ then write \begin{align} u-2 &= w_1r, & w &= w_1w_2, \\ u+2 &= w_2s, & 729w^3 + 972w^2 + 432w + 76 &= rs, \end{align} for integers $r,s,w_1,w_2$, and start slogging away from there. This method has worked well for me in the past, resulting in the discovery of elementary proofs of results which previously had only non-elementary proofs. But I feel like there must be a more elegant (but still elementary) method of attack from this point, so I'm looking for suggestions.

Kieren MacMillan
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1 Answers1

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Unless $|z|$ is very small, for $z >0,$ $z^4 - 6 z^2 + 4z - 3$ is larger than $(z^2 -3)^2$ but smaller than $(z^2 -2)^2.$ Since it is in between consecutive squares, it is not a square. For $z<0,$ it is between $(z^2 -4)^2$ and $(z^2 -3)^2.$

Will Jagy
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