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.
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...