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This question is a bit involved so thanks for your patience. After a lot of work I could show that each $b$ that satisfies this equation corresponds to another smaller integer say $s$ which satisfies the equation. here are the formulas for the solutions. $$t=s\pm\sqrt{2s^2+1}\tag{1}$$

$$x=3t\pm2\sqrt{2t^2-1}\tag{2}$$ $$y=\pm\sqrt{tx-1}\tag{3}$$ $$b=2x+y\tag{4}$$ $$a=3x+y\tag{5}$$ for nonzero $t$ and $x$. Let $x_+$ and $x_-$ denote the $x$ with the positive sign and negative sign respectively and define $y_+$ and $y_-$ similarly. Use the following rules: If you choose positive $t$ not $1$ you may choose any one of the $x$s, but you must choose $y_+$ if you take $x_+$. If $t=1$ you may take both $x$s and with $x_+$ you must choose only $y_+$. If you take $t$ positive and choose $x_-$ you must choose $y_-$, this last solution repeats the previous solutions but it generates one the others don't when $s=2$ , $t=5$ , $x_-=1$ , $y_-=-2$ , $b=0$ , $a=1$. There are similar rules for negative $t$ but they are just the negatives of the previous solutions. I (think) I could prove that no $s$ can lead to a $b=s$ since all solutions either get smaller or larger but the absolute value always increases except for the case I singled out which leads to $b=0$, which implies that this is the complete set of solutions. I noticed that the values of $b$ increase and at some point they reach a constant ratio which is roughly $\phi=93222358/15994428$ and I used this to derive the function $$f(x)=\phi^{\left(x-11\right)}93222358$$ which is extremely accurate at first and almost exact for $x\ge7$. It gives the $x$-th nonzero value of $b$ where the zeroth value is just $0$. Here's where it gets interesting, I noticed a pattern among the $a$s, that for nonzero even $x$, $a$ can be written in the form $a=c_1^2-1$ and for odd $x$, $a=\ c_2^2+1$. I checked this until $x=15$ and it holds. Also, another reason to believe that $a$ can't be a square is that while I was trying to prove Fermat's last theorem for $n=4$ (where this equation came up), I could reduce the proof to verifying that that no such $a$ can be a square. Of course, I'm a lot more interested now than when I first encountered the equation and I want to prove this pattern.

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    $\begingroup$ $\phi$ is most probably $(1 + \sqrt2)^2$. See also en.wikipedia.org/wiki/Pell%27s_equation. $\endgroup$ – lhf Sep 2 at 11:58
  • $\begingroup$ How is the formula $p_n=2p_{n-1}+p_{n-2}$ for the Pell numbers derived? $\endgroup$ – Km356 Sep 2 at 12:16
  • $\begingroup$ never mind I thought these were the solutions, but I just realized some of them aren't. $\endgroup$ – Km356 Sep 2 at 12:26
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I think there is a MUCH more elementary way to solve this.

Note that $a$ must be odd. So $a=2n+1$ for some integer $n$.

Solving $(2n+1)^2 = 1+2b^2$ gives $(2n^2+2n) = b^2$ which gives $2(n+1)n$ a perfect square.

This implies that if $n$ is even, then $n+1$ must be an odd square [indeed suppose $p^e$ divides $n+1$ for some odd prime $p$ and some positive integer $e$, then as $p$ does not divide $2(n+1)$ it follows that $e$ must be even] and thus $2n$ an even square $c^2$. So here $a=c^2+1$. And if $n$ is odd then $2(n+1)$ must be an even square $c^2$. So here $a=(2n+2) -1$ which is $c^2-1$.

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$a^2=1+2b^2$ is a Pell equation, $a^2-2b^2=1$.

The fundamental unit is $1+\sqrt2$ and it satisfies $a^2-2b^2=-1$.

Therefore, the solutions of $a^2-2b^2=1$ correspond to the powers of $(1+\sqrt2)^2=3+2\sqrt2$: $$ a_n + b_n \sqrt2 = (1+\sqrt2)^{2n} $$ and so are given by $$ a_{n+1}=3a_n+4b_n , \quad a_0 = 1 \\ b_{n+1}=2a_n+3b_n , \quad b_0 = 0 $$ and also by $$ a_{n+2}=6a_{n+1}-a_n , \quad a_0 = 1 , \quad a_1 = 3 \\ b_{n+2}=6b_{n+1}-b_n , \quad b_0 = 0 , \quad b_1 = 2 $$ because the minimal polynomial of $3+2\sqrt2$ is $x^2-6x+1$. Therefore, both ratios $a_{n+1}/a_n$ and $b_{n+1}/b_n$ converge to $3+2\sqrt2$.

Finally, we have $$ a_n + b_n \sqrt2 = (1+\sqrt2)^{2n} \\ a_n - b_n \sqrt2 = (1-\sqrt2)^{2n} $$ and so $$ 2a_n = (1+\sqrt2)^{2n} + (1-\sqrt2)^{2n} $$

Write also $$ u_n + v_n \sqrt2 = (1+\sqrt2)^{n} \\ u_n - v_n \sqrt2 = (1-\sqrt2)^{n} $$ Then $$ 2u_n = (1+\sqrt2)^{n} + (1-\sqrt2)^{n} $$ and so $$ 4u_n^2 = ((1+\sqrt2)^{n} + (1-\sqrt2)^{n})^2 = (1+\sqrt2)^{2n} + (1-\sqrt2)^{2n} + 2((1+\sqrt2)^{n}(1-\sqrt2)^{n}) \\ = 2a_n + 2(-1)^n $$ Since $u_n^2-2v_n^2=(-1)^n$, this gives $$ a_n = 2u_n^2 - (-1)^n = 2(2v_n^2+(-1)^n) - (-1)^n = (2v_n)^2+(-1)^n $$ as required. Note that we get that $c$ is even.

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  • $\begingroup$ Sorry to bother but I'm not familiar with these concepts. What is the fundamental unit and what do you mean by it satisfies this equation. Also how does this imply that the solutions are powers of this unit? I don't even understand the last line, I tried using the binomial theorem but couldn't see how it implies the last statement. I can see abstract algebra here since you mention the terms "minimal polynomial" and "unit", but I don't see the connection (my level is a first course in abstract algebra). $\endgroup$ – Km356 Sep 2 at 14:58
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If $a^2=2b^2+1$, then $$ (a-1)(a+1)=2b^2 $$ Note that $\gcd(a-1,a+1)\mid2$.

Since $2\mid(a-1)(a+1)$, we must have either $a\equiv1\pmod{4}$ or $a\equiv3\pmod{4}$.

If $a\equiv1\pmod{4}$, $\frac{a+1}2$ is odd and $\gcd\!\left(a-1,\frac{a+1}2\right)=1$. Thus, because $$ (a-1)\frac{a+1}2=b^2 $$ we have that $a-1=c^2$ and $\frac{a+1}2=d^2$. Therefore, $a=c^2+1$.

If $a\equiv3\pmod{4}$, $\frac{a-1}2$ is odd and $\gcd\!\left(\frac{a-1}2,a+1\right)=1$. Thus, because $$ \frac{a-1}2(a+1)=b^2 $$ we have that $\frac{a-1}2=d^2$ and $a+1=c^2$. Therefore, $a=c^2-1$.

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