I was wondering if it is possible to find out all $n \geq 1$, such that $3n^2+2n$ has an integral square root, that is, there exists $a \in \mathbb{N}$ such that $3n^2+2n = a^2$
Also, similarly for $(n+1)(3n+1)$.
Thanks for any help!
I was wondering if it is possible to find out all $n \geq 1$, such that $3n^2+2n$ has an integral square root, that is, there exists $a \in \mathbb{N}$ such that $3n^2+2n = a^2$
Also, similarly for $(n+1)(3n+1)$.
Thanks for any help!
$3n^2+2n=a^2$, $9n^2+6n=3a^2$, $9n^2+6n+1=3a^2+1$, $(3n+1)^2=3a^2+1$, $u^2=3a^2+1$ (where $u=3n+1$), $u^2-3a^2=1$, and that's an instance of Pell's equation, and you'll find tons of information on solving those in intro Number Theory textbooks, and on the web, probably even here on m.se.
Try similar manipulations for your other question.