Positive integer $n$ such that $2n+1$ , $3n+1$ are both perfect squares How many positive integer $n$ are there  such that $2n+1$ , $3n+1$ are both perfect squares ? 
$n=40$ is a solution . Is this the only solution ? Is it possible to tell whether finitely many or infinitely many solutions exist ? 
 A: If $2n+1=x^2$ and $3n+1=y^2$ then
$$3x^2-2y^2=1\ .$$
Multiplying by $-2$ and substituting $X=2y$, $Y=x$, this can be written as a Pell-type equation
$$X^2-6Y^2=-2\ .$$
This has infinitely many solutions, some of which are given by $X=X_n$, $Y=Y_n$ where
$$X_n+Y_n\sqrt6=(2+\sqrt6)(5+2\sqrt6)^n\ .\tag{$*$}$$
For example, taking $n=1$ gives
$$X=22,\ Y=9,\ x=9,\ y=11$$
and hence $n=40$, the solution you have already.  Equation $(*)$ gives the recurrences
$$X_{n+1}=5X_n+12Y_n\ ,\quad Y_{n+1}=2x_n+5y_n\ ,$$
and it is then possible to eliminate the $Y$ terms to get
$$X_{n+2}=10X_{n+1}-X_n$$
and similar relations for $x_n$ and $y_n$.
For a more detailed explanation of the method (applied to a slightly different equation), see my answer to this question.
A: The quick version is $n_0 = 0, \; \; n_1 = 40,$ then
$$ \color{magenta}{ n_{k+2} = 98 n_{k+1} - n_k + 40}.  $$
Given an $(x,y)$ pair with $3x^2 - 2 y^2 = 1$  we then take $n = (x^2-1)/ 2 = (y^2 - 1)/ 3. $
The first few $x,y$ pairs are
$$ x=1, \; y= 1 , \; n=0 $$
$$  x=9, \; y=11, \; n=40 $$
$$ x= 89,  \; y=109, \; n=3960  $$
$$  x=881, \; y=1079, \; n= 388080  $$
$$  x=8721, \; y=10681, \; n= 38027920  $$
$$  x=86329, \; y=105731, \; n= 3726348120  $$
and these continue forever with
$$  x_{k+2} = 10 x_{k+1} - x_k,  $$
$$  y_{k+2} = 10 y_{k+1} - y_k.  $$
$$  n_{k+2} = 98 n_{k+1} - n_k + 40.  $$
People seem to like these recurrences in one variable. The underlying two-variable recurrence in the pair $(x,y)$ can be abbreviated as
$$  (x,y) \; \; \rightarrow \; \; (5x+4y,6x+5y)  $$ beginning with
$$ (x,y) = (1,1)  $$
The two-term recurrences for $x$ and $y$ are just Cayley-Hamilton applied to the matrix
$$ A \; = \;  
 \left(  \begin{array}{rr}
  5  &  4  \\
   6   &  5  
\end{array} 
  \right)  ,
  $$
that being
$$ A^2 - 10 A + I = 0.   $$
A: If $2n+1$ is a square then it is (obviously?) of the form $4m^2+4m+1$ and thus $3n+1=6m^2+6m+1$ and so the question can be rephrased:

When is $6m^2+6m+1$ a square for integer $m$?

Which is trivially rephrased:

What are the integer solutions of $6x^2-y^2+6x+1=0$?

which can be answered at
http://www.alpertron.com.ar/QUAD.HTM
(sorry for the cop-out but it's better than nothing).
A: I don't think it is possible to explicitly find all such $n's$. This condition of both $2n+1, 3n+1$ being squares has appeared in a lot of contests such as the Putnam but the question asked is always to prove some implication from this condition. 
For example, one can prove that if $2n+1, 3n+1$ are squares, $5n+3$ cannot be a prime and $40|n$. 
A: We can do this modulo $4$. Since $n \equiv 0,1,2,3 \pmod{4}$, therefore
$$2n+1 \equiv 1,3,1,3 \pmod{4}$$ and
$$3n+1 \equiv 1,0,3,2 \pmod{4}.$$
However a square of an integer is only $0,1 \pmod{4}$. This means for both $2n+1$ and $3n+1$ to be squares $n \equiv 0 \pmod{4}$.
So let $n=4k$. Then we want $2n+1=8k+1$ and $3n+1=12k+1$ to be perfect squares. Let $8k+1=a^2$ and $12k+1=b^2$. Then 
$$4k=b^2-a^2=(b-a)(b+a).$$
Can you proceed from here? 
A: $2n+1=x^2$ and $3n+1=y^2$
since , $x $ is odd let $x=2m+1$
$2n+1=4m^2+4m+1$
$n=2 (m)(m+1) $...... (eqn1)
This means $4|n $; 
$y^2= 6m (m+1) $
$let y=2t+1$
From last two,
$3 (m)(m+1)=2t (t+1) $
Since , $2|t (t+1) $; this implies  $4|m (m+1) $ and this implies $8|n $ ... (from eqn 1).
Now it remains to prove that $5|n$
Qudratic residue for $mod5$ are {0,1,4}
So, $x^2= {0 or 1 or 4}$ $ mod5$
Also $y^2= {0 or 1 or 4}$ $ mod 5$
The only possible is $ n= 0 mod 5$. (According to first two equations)
