<s>[INST] if $4n^2-1=3m^2$ has a positive integer solution, show that $2n-1$ always square Let $n\in \mathbb{N}$, if $4n^2-1=3m^2$ has a positive integer solution, show that $2n-1$ is a perfect square.
For example $n=1$, it is clear.
and $n=13$ because  $$ 4\cdot 13^2-1=3\cdot 15^2$$ and $$2n-1=25=5^2$$
 [/INST] An Experimental approach:
There are infinitely many integers like n such that:
$2n-1=k^2$; $k∈N$
They make following series:
$n=13, 41, 61, 181, 221 . . .$ 
therefore we may write:
$4\times 13^2-1=3\times 15^2$
$4\times 41^2-1=3\times 47.34^2$
$4\times 61^2-1=3\times 70.4^2$
$4\times 181^2-1=3\times 209^2$
$4\times 221^2-1=3\times 255.19^2$
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For $4 . n^2-1=3 .m^2$ The condition $2n-1=k^2$ is necessary but not sufficient. That is if  equation ($4 . n^2-1=3 .m^2$) has integer solution then ($2n-1$) is definitely is a square. 
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