Find all $n$ for which $3n^2+3n+1$ is a perfect square. Find all natural numbers $n$ for which $3n^2+3n+1$ is a perfect square.
I used discriminant method but failed. Then I found upper and lower bounds of this expression:
Lower:$(n+1)^2$
Upper:$(2n)^2$
But, this too does not seem to be useful. Please help me.
 A: Cool.  I'll just fully answer the question.
Starting with $ \ m^2=3n^2+3n+1\iff (2m)^2-3(2n+1)^2=1 \quad $.  Lets call  $ \ p=2m, \ \ $ $q=2n+1 \ \ $.  So it's now $$p^2-3q^2=1$$
By a quick inspection the smallest solution is:  $(p_0,q_0)=(2,1) \quad $.  It can be used to find all others.  Use it to write the number $1$ in a funny/arbitrary way, and multiplying our equation with this special $1$ will lead to the rest of the solutions.  The back-substitution, $(p,q,p_0,q_0) \to (2m,2n+1,2,1)$, will wait until the end to elucidate how this algorithm can be applied to other situations involving pell-type equations:
$$
\begin{align} 
p_0^2-3q_0^2&=1\\
(p_0-q_0 \sqrt 3)(p_0+q_0 \sqrt 3)&=1\\
(p_0-q_0 \sqrt 3)^2(p_0+q_0 \sqrt 3)^2&=1^2=1\\
\bigg[(p_0^2+3q_0^2)-(2p_0q_0) \sqrt 3\bigg]\bigg[(p_0^2+3q_0^2)+(2p_0q_0)\sqrt 3\bigg]&=1\\
\text{now multiply $p^2-3q^2=1$ by this "$1$" in the following way (factor it first):}  &  \\
(p-q \sqrt 3) \cdot \bigg[(p_0^2+3q_0^2)-(2p_0q_0) \sqrt 3\bigg]& \cdot \\
(p+q \sqrt 3) \cdot \bigg[(p_0^2+3q_0^2)+(2p_0q_0) \sqrt 3\bigg]&=1\\
&\vdots \\
\underbrace{\bigg[(p_0^2+3q_0^2)p+(2\cdot 3p_0q_0)q\bigg]^2-3\bigg[(2p_0q_0)p +(p_0^2+3q_0^2)q\bigg]^2=1}_{=p^2-3q^2=1}\\
\end{align}
$$
Interpretable as
$$(p_k,q_k) \xrightarrow{k \to k+1} \bigg((p_0^2+3q_0^2)p+(2\cdot 3p_0q_0)q \ \ , \ (2p_0q_0)p +(p_0^2+3q_0^2)q\bigg)$$
Evaluating
$$
\begin{cases}
p_k&=2m_k\\
p_0=2 \implies m_0&=1\\
q_k&=2n_k+1 \\
q_0=1 \implies n_0&=0\\
\end{cases}
$$
Thus
$$(2m_k,2n_k+1) \xrightarrow{k \to k+1} \bigg( (7)(2m_k)+(12)(2n_k+1) \ \ , \ (4)(2m_k) +(7)(2n_k+1)\bigg)$$
Or, finally,
$$(m_k,n_k) \xrightarrow{k \to k+1} \bigg( 7m_k+12n_k+6 \ \ , \ 4m_k +7n_k+3\bigg)$$
Which is exactly @S.Dolan's ordered pair.
Also expressable as
$$
\begin{pmatrix} 
m_k \\
n_k 
\end{pmatrix}
\xrightarrow{T}
\begin{pmatrix}
7  & 12 \\
4  & 7 
\end{pmatrix}
\cdot 
\begin{pmatrix} 
m_k \\
n_k 
\end{pmatrix}
+
\begin{pmatrix}
6 \\
3
\end{pmatrix}
$$
if you're into that sort of thing...
A: Using @J.G. 's observation, all solutions are generated by the following procedure.
From any solution e.g. $n=0,m=1$, one obtains a further solution as
$$(7n+4m+3,12n+7m+6).$$
