Infinitely many squares of the form $3a^2 + 1$

Question

I am working on the following proof.

Prove that $3a^2 + 1$ (as $a$ ranges over the integers) produces infinitely many squares.

Proof:

We first note that all square numbers can be represented as $3k, 3k+1, 3k+2$.

Cases:

(i) $3a^2+1=3k \rightarrow a^2-k=\frac{-1}{3} \rightarrow 3 \nmid -1$, thus we conclude that $3a^2+1$ will not produce squares of the form $3k$.

(ii) $3a^2+1=3k+1 \rightarrow a^2-k=0 \rightarrow 3 \mid 0$, thus we conclude that $3a^2+1$ will produce squares of the form $3k+1$ given $a^2=k$.

(iii) $3a^2+1=3k+2 \rightarrow a^2-k=\frac{1}{3} \rightarrow 3 \nmid 1$, thus we conclude that $3a^2+1$ will not produce squares of the form $3k+2$.

Case (ii) is the only case where $3a^2+1$ will produce a square number. I think I should be able to make the conclusion that since $a^2=k$, $3a^2+1$ will produce infinitely many squares, but I am not confident on the conclusion.

I examined a couple of numbers and found that:

$a=4 \rightarrow k=16=4^2$

$a=15 \rightarrow k=225=15^2$

I am not positive on where to go from here, a hint would be helpful. Thank you.

Answer 1

$$q^2-3a^2 = 1 $$ is a Pell's equation with infinitely many solutions, since $q^2-3a^2$ is the norm of $q+a\sqrt{3}$ in $\mathbb{Z}[\sqrt{3}]$ and $2+\sqrt{3}$ has unit norm, so $(2+\sqrt{3})^n$ also has unit norm. If we start with the fundamental solutions $(q_0,a_0)=(1,0)$ and $(q_1,a_1)=(2,1)$ it follows that the sequences $$ \{q_n\}=\{1,2,7,26,\ldots\},\qquad \{a_n\}=\{0,1,4,15,\ldots\} $$ with characteristic polynomial $x^2-4x+1$ give an infinite number of solutions $(q_n,a_n)$.

Answer 2

Suppose there is one solution $x^2=3a^2+1$. Then $(x^2-3a^2)(x^2-3a^2)=1 \implies$ $$(x^2+3a^2)^2-3(2ax)^2=1$$ is another solution when you set

$x'=x^2+3a^2$

$a'=2ax$

Note that $x'>x$ and $a'>a$ for $a, x>0$

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To see at least one solution exists, just take $x=2, a=1$.