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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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1 Answer 1

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First, note that the original equation has the solution $a=0,b=0$.

Now multiply through by $k$, and let $u=k(2a+k+1),v=2b+k$ to get $$u^2-k(k+1)v^2=k^2(k+1)\tag1$$ Again, this has the solution $u=k^2+k,v=k$. Now consider the Pellian $$u^2-k(k+1)v^2=1\tag2$$ Since $k(k+1)$ is not a square, this Pellian has infinitely many solutions. Do you know how to combine a solution of (2) with the solution we have for (1) to get another solution for (1)? If so, you now have infinitely many solutions for (1). Then see if you can work your way back from $u,v$ to $a,b$.

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  • $\begingroup$ Ok, but how can I show that all the solutions will have: $$ u \equiv k+1 \pmod 2 $$ and $$ v \equiv k \pmod 2 $$? $\endgroup$ Commented Jun 22, 2012 at 7:26
  • $\begingroup$ Why would you want to do that? From $u=k(2a+k+1)$ it's clear that $u$ is even, and from $v=2b+k$ it's clear that $v\equiv k\pmod2$. $\endgroup$ Commented Jun 22, 2012 at 7:37
  • $\begingroup$ Sorry, I wasn't clear enough. How can I work my way back from $u, v$ to $a, b$? $\endgroup$ Commented Jun 22, 2012 at 7:58
  • $\begingroup$ Can you not solve $u=k(2a+k+1)$ for $a$, and $v=2b+k$ for $b$? $\endgroup$ Commented Jun 22, 2012 at 9:12

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