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

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.

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

Hint: $$i^2-29^2=j^2\implies (i-j)(i+j)= 29^2$$
$$\begin{array}{c|c} i+j & i-j & 2i& i \\ \hline 1& 29^2 & 1+29^2 &421\\ 29&29&58&29\\ 29^2&1& 1+29^2& 421\\ -1& -29^2 & -1-29^2 &-421\\ -29&-29&-58&-29\\ -29^2&-1& -1-29^2&-421 \end{array}$$
• $29^2$ 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 $j$. – BalancedTryteOperators Feb 22 at 10:03