prove that :

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Blazing goIITian

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ok,

for nequiv0mod 5

2n+1,3n+1equiv1mod 5 so it is possible.

for nequiv1mod5

2n+1equiv3mod5 not possible...

for

$nequiv2mod5\\Rightarrow 2n+1equiv0mod5( ext{possible})\\And\\3n+1equiv2mod5( ext{not possible})$

For

for

So, we can say that for 2n+1 and 3n+1 to be squares n must be divisible by 5..

[IMG]http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/0/6/3/0638cc54ee34d87aa803d78fe3ec072aa8749bf5.gif[/IMG] :)

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