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I neeed to prove that this equation: $$ k(2a+(1+k))^2-(k+1)(2b+k)^2=k(k+1) $$ has infinite solution for $a,b \in \mathbb{N}^+$ and $k$ constant (positive integer too).

but I have a sort of hint (doesn't really help me):

If $k$ is odd I can substitute $kc$ to $b$, obtaining $$ (2a+k+1)^2-(k(k+1))(2c+1)^2=(k+1) $$ And apparently I can do some trick like this if $k+1$ is odd, subtituting $a$ with something. But I can't reach anything like Pell's.

Then I can prove with Pell's that $$m^2-(k(k+1))n^2=k+1$$ has infinite positive integer solutions for $m$ and $n$, and the hint now says that it is done. But I wouldn't say so, because if in all the solutions $n$ turns out to be even $2c+1 \neq n$ and therefore there are no solution fo the original equation?

Thank you!

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