The recursive sequence defined recursive $a_{n+2} = 6a_{n+1} - a_n, n \in \mathbb{N}$, with $a_ 1=1$ and $a_2 = 7$ can be explicitly expressed as $a_n = {\textstyle \frac{(\sqrt{2} + 1)^{2n-1} -(\sqrt{2} - 1)^{2n-1}}{2}}$. Hence we obtain $(1) \;\; \sqrt{2a_n^2 + 2} = \sqrt{\Big[{\textstyle \frac{(\sqrt{2} + 1)^{2n-1} - (\sqrt{2} - 1)^{2n-1}}{2}} \Big]^2 + 2} = {\textstyle \frac{(\sqrt{2} + 1)^{2n-1} + (\sqrt{2} - 1)^{2n-1}}{\sqrt{2}}} = 2b_n$ From the theory of numbers we know that $y = b_n$ indeed satisfies the Pell equation $x^2 - 2y^2 = -1$. Therefore $b_n$ is a positive integer which according to (1) satisfies $(2) \;\; 2b_n^2 = a_n^2 + 1$. Next assume there is a positive integer $m$ s.t. $a_n = 2m^2 - 1$, which inserted in (2) result in $b_n^2 = 2m^4 - 2 m^2 + 1$, i.e. $(m^2 - 1)^2 + (m^2)^2 = b_n^2$. Hence $(m^2 - 1, m^2, b_n)$ is a primitive Pythagorian triple, implying there is two coprime positive integers $s > t$ s.t. $s-t$ is odd and $(3) \;\; m^2 - p = s^2 - t^2$, $(4) \;\; m^2 - q = 2st$, where $(p,q) = (1,0)$ or $(p,q) = (0,1)$. Let us consider these two cases: Case 1: $(p,q) = (1,0)$. Then by (3) and (4) $(5) \;\; m^2 - 1 = s^2 - t^2$, $(6) \;\; m^2 = 2st$. Now $m$ is even by (6), yielding $s$ is even by (5). Hence, since $GCD(s,t)=1$, we obtain $s=2u^2$ and $t=v^2$ for two positive coprime integers $u$ and $v$. Subtracting (5) from (6), the result is $1 = -s^2 + 2st + t^2 = -(2u^2)^2 + 4u^2v^2 + (v^2)^2$, i.e. $(7) \;\; (v^2 + 2u^2)^2 = 8u^4 + 1$. It is well known that the only positive solution of the Diophantine equation $y^2 = 8x^4 + 1$ is $(x,y) = (1,3)$. Hence $v^2 + 2u^2 = 3$ by (7), yielding $u=v=1$, which means $m = 2uv = 2 \cdot 1 \cdot 1 = 2$ and $a_n = 2m^2 - 1 = 2 \cdot 2^2 - 1 = 8 - 1 = 7 = a_2$. Case 2: $(p,q) = (0,1)$. Then by (3) and (4) $(8) \;\; m^2 = s^2 - t^2 = (s - t)(s + t)$, $(9) \;\; m^2 - 1 = 2st$. Now, since $GCD(s,t)=1$ and $s-t$ is odd, $GCD(s-t,s+t) = 1$, which according to (8) means there are two odd coprime positive integers $c$ and $d$ s.t. $s+t = c^2$ and $s-t = d^2$. Hence $2s = c^2 + d^2$ and $2t = c^2 - d^2$, which inserted in (8) and (9) result in $m=cd$ and $2(m^2 - 1) = (2s) \cdot (2t) = (c^2 + d^2)(c^2 - d^2) = c^4 - d^4 = 2((cd)^2 - 1)$. Consequently $(c^2 - d^2)^2 - 2d^4 = -2$, i.e. $(10) \;\; d^4 - 1 = {\textstyle 2 \Big[ \frac{c^2 - d^2}{2} \Big]^2}$. A theorem of Cohn states that the only possible positive solutions of the Diophantine equation $x^4 - dy^2 = 1$ is given by $x^2 = a^2$ or $x^2 = 2a^2 - 1$, where $a + b\sqrt{d}$ is the fundamental solution of the Pell equation $A^2 - dB^2 = 1$. Hence with $d=2$ we obtain $a + b\sqrt{2} = 3 + 2\sqrt{2}$, which give us $x^2 = 3$ and $x^2 = 2 \cdot 3^2 - 1 = 2 \cdot 9 - 1 = 18 - 1 = 17$. Therefore $x^4 - 2y^2 = 1$ has no positive solutions, implying $c=d=1$ by (10). Hence $m = cd = 1 \cdot 1 = 1$, meaning $a_n = 2m^2 - 1 = 2 \cdot 1^2 - 1 = 2 \cdot 1 - 1 = 2 - 1 = 1 = a_1$. Conclusion: The only positive integers for which $a_n$, is of the form $2m^2-1$ is $n=1$ and $n=2$.

For diophantine (10) $x^4-1=2y^2$ $x,y>0$ (and without Cohn's theorem) so $(x^2-1)(x^2+1)=2y^2$ , $(x^2-1,x^2+1)=2$ so $x^2-1=4k^2$ , $x^2+1=2l^2$ , $(k,l)=1$ so $x=1$

For $x^4-1=2y^2$ we have $x^4+y^4=(y^2+1)^2, x,y > 0$, which has no solutions (p.34) if $xy \ne 0$. About the equation $y^2=8x^4+1$ has only one positive solution $(1,3)$, how do you prove such result? I developed this to the equation $b^4-2a^4=1$, but I couldn't solve it.

Seventh wrote: About the equation $y^2=8x^4+1$ has only one positive solution $(1,3)$, how do you prove such result? I developed this to the equation $b^4-2a^4=1$, but I couldn't solve it. I think you already knew how to solve $b^4-2a^4=1$. Seventh wrote: For $x^4-1=2y^2$ we have $x^4+y^4=(y^2+1)^2, x,y > 0$, which has no solutions (p.34) if $xy \ne 0$.

My mistake, I wrote the wrong equation. I meant $b^4-2a^4=-1$ instead $b^4-2a^4=1$. Here is what I've got: $y^2=8x^4+1 \iff y^2+(4x^4)^2=(4x^4+1)^2$, so $4x^4+1=m^2+n^2, 4x^4=2mn \Rightarrow (m-n)^2=1$. WLOG $m \ge n\rightarrow 4x^4=2n(n+1) \iff 2x^4=n(n+1)$. If $n$ is even, we have $n=2a^4, n+1=b^4$, thus $b^4=2a^4+1 \Rightarrow b^4+a^8=(a^4+1)^2$, which has no solutions if $ab \ne 0$. If $n$ is odd, we have $n=b^4, n+1=2a^4$, where $b$ is odd, thus $b^4-2a^4=-1$, but I couldn't solve it.

Seventh wrote: My mistake, I wrote the wrong equation. I meant $b^4-2a^4=-1$ instead $b^4-2a^4=1$. Here is what I've got: $y^2=8x^4+1 \iff y^2+(4x^4)^2=(4x^4+1)^2$, so $4x^4+1=m^2+n^2, 4x^4=2mn \Rightarrow (m-n)^2=1$. WLOG $m \ge n\rightarrow 4x^4=2n(n+1) \iff 2x^4=n(n+1)$. If $n$ is even, we have $n=2a^4, n+1=b^4$, thus $b^4=2a^4+1 \Rightarrow b^4+a^8=(a^4+1)^2$, which has no solutions if $ab \ne 0$. If $n$ is odd, we have $n=b^4, n+1=2a^4$, where $b$ is odd, thus $b^4-2a^4=-1$, but I couldn't solve it. For positive integers... $y^2=8x^4+1$ so $y$ odd so set $y=2z+1$ so $z(z+1)=2x^4$ Case 1 $z=k^4$ , $z+1=2l^4$ so $2l^4=k^4+1$ so $(\dfrac{k^4-1}{2})^2=(l^2)^4-k^4$ so $z=1$ or $z=1$ and finally $(y,x)=(3,1)$ Case 2 $z=2k^4$ , $z+1=l^4$ so $l^4-1=2k^4$ so there are $a,b$ , $l^2-1=16a^4$ , $l^2+1=2b^4$ so $a=0$ , $z=0$ so no positive integer solutions...

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