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Determine all integers $i$ such that $$(i-29)(i+29)$$ is a square number.

I’ve tried some substitutions but none of them worked... I think that the only solutions are $i=\pm 29$, but I still don’t know how to prove it.

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    $\begingroup$ How about $i=421$ ... $\endgroup$ – Matti P. Feb 22 at 9:43
  • $\begingroup$ @MattiP. Related to Pythagorean triple $421^2 = 420^2+29^2$. Any odd positive number is the difference between consecutive squares, for $2N+1=(N+1)^2-N^2$. $\endgroup$ – Jeppe Stig Nielsen Aug 3 at 12:35
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Hint: $$i^2-29^2=j^2\implies (i-j)(i+j)= 29^2$$

$$ \begin{array}{c|c} i+j & i-j & 2i& i \\ \hline 1& 29^2 & 1+29^2 &421\\ 29&29&58&29\\ 29^2&1& 1+29^2& 421\\ -1& -29^2 & -1-29^2 &-421\\ -29&-29&-58&-29\\ -29^2&-1& -1-29^2&-421 \end{array} $$

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  • $\begingroup$ $29^2$ is a semiprime number. There are only six ways to decompose it into two integer parts. Only the first two columns of the table are relevant to the decomposition of $29^2$. The last two are finding $i$ so you can find $j$. $\endgroup$ – BalancedTryteOperators Feb 22 at 10:03

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