I'm trying to show, that the solutions $t_i$ from Pell's equation $t^2-2s^2=\pm 1$ are not squares (only for the trivial cases). For the even solutions $t_{2i}$ (solutions for the + sign) this is easy, but for the odd solutions $t_{2i+1}$ (solutions for the - sign) this is a little bit harder.
This question leads to the diophantine equation $\tau^4-2s^2=-1$ and the new question, whether it has nontrivial solutions or not.
I do the following approach. First of all I take the equation $\tau^4+1=u^2+v^2$ (*), try to find solutions $(u, v, \tau)$ and set in the end $s=u=v$. With
$\begin{align} u&=u_2 a^2 +u_1 a+u_0 \\ v&=v_2 a^2 +v_1 a+v_0 \\ \tau&=\tau_1 a+\tau_0 \end{align}$
and comparison of the coefficients one gets
$\begin{align} u&=2a^2 b^2 (b^4 +1)+4ab^3 +1 \\ v&=a^2 (b^8-1)+2ab(b^4 -1)+b^2 \\ \tau&=a(b^4 +1)+b \end{align}$
with natural numbers $a$ and $b$.
Now the question: are there any other solutions for (*) and if not, why? Maybe there is also an easy way showing, that the $t_{2i+1}$ are not squares, then I would prefer this proof.
thank you
I'll assume your parametrization attempts to cover all solutions up to sign, and up to switching $u$ and $v$, because in your parametrization clearly $u$ and $v$ are both nonnegative, and $u$ is odd. Even then the parametrization is incomplete; there are no natural numbers $a$ and $b$ corresponding to the solution $(\tau,u,v)=(8,31,56)$.
Here's the first approach that springs to my mind; it's not necessarily much easier, and the final step is a bit unsatisfying, but it's certainly too much for a comment:
If $\tau$ and $s$ are integers such that $\tau^4-2s^2=-1$ then clearly $\tau$ is odd, and $$-\tau^4=1-2s^2=(1+s\sqrt{2})(1-s\sqrt{2}).\tag{1}$$ This suggests an argument in $\Bbb{Z}[\sqrt{2}]$, which is a UFD with fundamental unit $1+\sqrt{2}$. Indeed the two factors on the right hand side are coprime, so both are fourth powers up to units. Then without loss of generality $1+s\sqrt{2}=(1+\sqrt{2})^kr^4$ with $k\in\{0,1,2,3\}$, and comparing norms in $(1)$ shows that $k=1,3$. This yields four quartic Thue equations, two of which have no solutions by reducing mod $4$. The other two have only the trivial solutions $s=\pm1$, found with the aid of a computer package.
Edited this section to provide excruciating detail and the absolutely minimal amount of theory:
If $\eta\in\Bbb{Z}[\sqrt{2}]$ divides both factors on the right hand side of $(1)$, then $\eta$ divides their sum, which is $$(1+s\sqrt{2})+(1-s\sqrt{2})=2.$$ Because $\Bbb{Z}[\sqrt{2}]$ is a unique factorization domain, and $2$ factors as $2=\sqrt{2}^2$ where $\sqrt{2}\in\Bbb{Z}[\sqrt{2}]$ is prime, it follows that $\eta=u\sqrt{2}^m$ for some unit $u\in\Bbb{Z}[\sqrt{2}]^{\times}$ and some $m\in\{0,1,2\}$. If $m\neq0$ then $\sqrt{2}$ divides both factors, and hence $-\tau^4$ is divisible by $\sqrt{2}^2=2$. But we already noted that $\tau$ is odd; a contradiction. Hence $m=0$ and so $\eta=u$ is a unit in $\Bbb{Z}[\sqrt{2}]$. This shows that the two factors on the right hand side of $(1)$ have no common prime factors. Factoring the two factors as $$1+s\sqrt{2}=v\cdot p_1^{e_1}\cdots p_k^{e_k} \qquad\text{ and }\qquad 1-s\sqrt{2}=w\cdot q_1^{f_1}\cdots q_l^{f_l},$$ for units $v,w\in\Bbb{Z}[\sqrt{2}]^{\times}$, pairwise coprime primes $p_i,q_j\in\Bbb{Z}[\sqrt{2}]$, and positive integers $e_i,f_j\in\Bbb{Z}_{>0}$, we see that $$\tau^4=-(1+st\sqrt{2})(1-s\sqrt{2})=-vw\cdot p_1^{e_1}\cdots p_k^{e_k}q_1^{f_1}\cdots q_l^{f_l}.$$ Because the left hand side is a fourth power, it follows that all $e_i$ and $f_j$ are multiples of $4$. That is to say, we have $$1+s\sqrt{2}=vr^4\qquad\text{ and }\qquad1-s\sqrt{2}=ws^4,$$ for some units $v,w\in\Bbb{Z}[\sqrt{2}]^{\times}$ and $r,s\in\Bbb{Z}[\sqrt{2}]$. The unit group of $\Bbb{Z}[\sqrt{2}]$ is generated by $-1$ and the fundamental unit $1+\sqrt{2}$, so $v=\pm(1+\sqrt{2})^m$ for some integer $m$, and of course $r=x+y\sqrt{2}$ for some integers $x,y\in\Bbb{Z}$, so we find that $$1+s\sqrt{2}=\pm(1+\sqrt{2})^m(x+y\sqrt{2})^4,$$ for some $m,x,y\in\Bbb{Z}$. Without loss of generality we may assume that $m\in\{0,1,2,3\}$, because $$(1+\sqrt{2})^m(x+y\sqrt{2})^4=(1+\sqrt{2})^{m-4n}\big((1+\sqrt{2})^n(x+y\sqrt{2})\big)^4,$$ so we can reduce $m$ modulo $4$ and adjust $x$ and $y$ accordingly. Because the map $$\Bbb{Z}[\sqrt{2}]\ \longrightarrow\ \Bbb{Z}[\sqrt{2}]:\ x+y\sqrt{2}\ \longmapsto\ x-y\sqrt{2},$$ is an isomorphism, it follows that $$1-s\sqrt{2}=\pm(1-\sqrt{2})^m(x-y\sqrt{2})^4,$$ and so \begin{eqnarray*} \tau^4&=&-(1+s\sqrt{2})(1-s\sqrt{2})\\ &=&-\big(\pm(1+\sqrt{2})^m(x+y\sqrt{2})^4\big)\big(\pm(1-\sqrt{2})^m(x-y\sqrt{2})^4\big)\\ &=&-\big((1+\sqrt{2})(1-\sqrt{2})\big)^m\big((x+y\sqrt{2})(x-y\sqrt{2})\big)^4\\ &=&-(-1)^m(x^2-2y^2)^4, \end{eqnarray*} where of course $\tau^4$ and $(x^2-2y^2)^4$ are positive, so $m$ is odd, i.e. $m=1$ or $m=3$. This shows that either $$1+s\sqrt{2}=\pm(1+\sqrt{2})(x+y\sqrt{2})^4 \qquad\text{ or }\qquad 1+s\sqrt{2}=\pm(7+5\sqrt{2})(x+y\sqrt{2})^4,$$ for some integers $x,y\in\Bbb{Z}$. Expanding the fourth power and comparing coefficients shows that \begn{eqnarray*} \pm1&=&x^4+8x^3y+12x^2y^2+16xy^3+4y^4,\ \pm1&=&7x^4+40x^3y+84x^2y^2+80xy^3+28y^4, \end{eqnarray*} and then reducing mod $4$ shows that we must have the $=$-sign in the first case, and the $-$isgn in the second case.emphasized text These are two Thue equations, for which there exist effective methods to determine all integral solutions. With the help of a computer I found that $(x,y)=\pm(1,-1)$ and($(x,y)=(\pm1,0)$ are the only integral solutions, and these all correspond to $s,\tau=\pm1$.