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Square and triangular numbers are expressed as

$n^2=\frac{m(m+1)}{2}$

Further on this can be expressed as

$8n^2=4m(m+1)=4m^2+4m=(2m+1)^2-1$

Taking $a=2m+1$ and $b=2n$ the expression becomes

$2b^2=a^2-1$ or $1=a^2-2b^2$

After a bit of factoring previous equation becomes

$1=(a-\sqrt{2}b)(a+\sqrt{2}b)$

One of the solutions is $(a,b)=(3,2)$ and $(m,n)=(1,1)$. From here additional solutions can be found recursively.

Once there is a solution say $(m,n)$ there is another $(1+im+jn, 1+km+ln)$ for some integers $i,j,k,l$. I need help proving this.

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  • $\begingroup$ Why do you say "previous equation has no integer solutions" when it clearly does (because you give the solution $a=3, b=2$ only two lines later)? $\endgroup$ – Gabriel Burns Oct 28 '16 at 17:46
  • $\begingroup$ @GabrielBurns actually just the polynomial can't be expressed in integer terms $\endgroup$ – php-- Oct 28 '16 at 17:50
  • $\begingroup$ It has something to do with "triangle" and "square" numbering of the elements of $mathbb N^2$. $\endgroup$ – hamam_Abdallah Oct 28 '16 at 17:54
  • $\begingroup$ I don't think there are other solutions since the "triangle" numbering is faster than the square . $\endgroup$ – hamam_Abdallah Oct 28 '16 at 17:56

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