For $|y|<2$ the solutions $(x,y)$ are $(0,-1),(0,1)$. Let's assume $y>1$, the equation is equivalent to $(y+1)(y-1)=2x^4$. Because $y+1,y-1$ are coprime for odd divisors, then there exist such odd, coprime numbers $u,v$ that $x=2^tuv$ and $y-1=2^mu^4$ , $y+1=2^nv^4$, where $m,n$ are postive integers and $t$ is a non-negative positive integer. Let's note that $m+n = 4t+1= 1$ $(\mod 4)$. So we have $2^nv^4-2^mu^4=2$, so $min\{m,n\}=1$ and $max\{m,n\}>min\{m,n\}$: 1) $m=1$ thus $2^{n-1}v^4-u^4=1$ ($n>1$), but because $m+n = 1$ $(\mod 4)$ so $n-1=3$ $(\mod 4)$,so $n \geq 3$. However, $u^4+1=2$ $(\mod 4)$, so we have a contradiction because $8|2^{n-1}$ so $2^{n-1}=0$ $(\mod 4)$. 2)$n=1$ thus $v^4-2^{m-1}u^4=1$ ($n>1$), so $(v^2-1)(v^2+1)=2^{m-1}u^4$. Because $v^2+1,v^2-1$ ($v>1$) are coprime for odd divisors, then there exist such odd, coprime numbers $a,b$ that $u=ab$ and $v^2-1=2^{m-2}a^4$ , $v^2+1=2b^4$. Thus $(v+2^{\frac{m-2}{2}}a^2)(v-2^{\frac{m-2}{2}}a^2)=1$ ( because $m+n=1$ $(\mod 4)$ and $n=1$ then $m-2$ is even), but that is impossible. 1) and 2) haven't given any solutions, so there aren't any solutions for $y>1$ and also for $y<-1$- the equality is symmetric, so the only solutions are: $(0,1),(0,-1)$.

if we don't consider trivial answers,then no solution, we know that $ x$ is even so let's replace $ x$ by $ 2x$ and we get, $ 32x^4 = (y - 1)(y + 1)$ so we have 2 cases : 1-$ y \equiv - 1 \pmod {4}$ which implies $ y - 1 = 2t^4,y + 1 = 16s^4,(t,s) = 1,ts = x$ so we get $ 16s^4 - 2t^4 = 2 \implies 8s^4 = t^4 + 1$ but we obviously have $ t^4 + 1 \equiv 2 \pmod {8}$. so this case is ommitted. 2-$ y \equiv 1 \pmod {4}$ which implies $ y - 1 = 16s^4,y + 1 = 2t^4,(t,s) = 1,ts = x$ so we get $ 2t^4 - 16s^4 = 2 \implies t^4 - 8s^4 = 1$. or in other words, $ t^4 - 1^4 = 2(2s^2)^2$. u can easily prove that the equation $ x^4 - y^4 = 2z^2$ has no integer solutions(by infinite descent). so we r done!

x^4 = 1 mod 5 2*x^4+1 = 3 mod 5 no y that y^2 = 3 mod 5

skytin wrote: x^4 = 1 mod 5 2*x^4+1 = 3 mod 5 no y that y^2 = 3 mod 5 what if $ 5 | x$ ?

We may first assume $ x,y \geq 0$. $ 2x^{4}=y^{2}-1=(y+1)(y-1)$. $ 2|(y+1)(y-1) \Rightarrow 4|(y+1)(y-1) \Rightarrow 2|x$. Let $ x=2x_{1}$, then $ 32x_{1}^{4}=(y+1)(y-1) \Rightarrow y=16y_{1}+1$ or $ 16y_{1}-1$. Then $ x_{1}^{4}=y_{1}(8y_{1}+1)$ or $ y_{1}(8y_{1}-1)$. But since $ gcd(y_{1},8y_{1}-1)=gcd(y_{1},8y_{1}+1)=1$, both $ y_{1}$ and $ 8y_{1}+1$ (or $ -1$) must be $ 4^{th}$ powers. But $ 8y_{1}-1$ cannot be a $ 4^{th}$ power, so we can only have $ x_{1}^{4}=y_{1}(8y_{1}+1)$ (i.e. $ y=16y_{1}+1$). Note that $ y_{1},8y_{1}+1$ are still $ 4^{th}$ powers, so let $ y_{1}=a^{4}$, $ 8y_{1}+1=b^{4}$. Then $ b^{4}-1=(b^{2}+1)(b+1)(b-1)=8a^{4}$. Clearly b is odd. Consider then the integers $ \frac{b^{2}+1}{2},\frac{b+1}{2},\frac{b-1}{2}$. Taken pairwise, each gcd is 1 ($ \frac{1}{2}(b+1)(b-1)=\frac{b^2-1}{2}=\frac{b^2+1}{2}-1$). However, their product = $ a^{4}$, so each of them must be a $ 4^{th}$ power. In particular, $ \frac{b+1}{2},\frac{b-1}{2}$, 2 consecutive integers, must both be $ 4^{th}$ powers, possible only when $ \frac{b-1}{2}=0, \frac{b+1}{2}=1 \Rightarrow b=1 \Rightarrow y=16y_{1}+1=2(8y_{1}+1)-1=2b^{4}-1=1$, and $ x=0$. $ \therefore (x,y)=(0,1),(0,-1)$ are the only solutions in integers.

im not sure if my ideas are correct cuz i wasnt sure how to proceed but this is what i had: $ 2x^4=(y-1)(y+1)$ we take this mod 4 and find that $ x$ is even and $ y=4k+1$ or $ y=4k+3$ with $ y=4k+3$, we find a contradiction. so, $ y=4k+1$ let $ x=2a$ $ 32a^4+1=(4k+1)^2$ then we have $ 8a^4=(2k)(2k+1)$ i think we could use infinite descent as mentioned before to show that either a or k must be 0 but im not sure how to. does anyone know how?

Here is my solution I will assume $y$ is positive. $2x^4+1=y^2\implies 2x^4=(y-1)(y+1)$, so replace $y-1$ with $2z$. This gives $2x^4=4z(z+1)$, and again we will replace $x$ with $2x_1$ to get $8x_1^4=z(z+1)$. Now $\gcd(z,z+1)=1$, hence either $z=8a^4$ and $z+1=b^4$ or that $z=a^4$ and $z+1=8b^4$ for some integers $a,b$ (since $y$ is positive- we can ignore the case $z=-8a^4$ and $z+1=-b^4$ etc). Case 1 $z=8a^4$ and $z+1=b^4$ Then $8a^4+1=b^4$ which rearranges to $8a^4=(b-1)(b+1)(b^2+1)$. Since $b$ must be odd, substitute $b-1=2c\implies b+1=2(c+1)$ and $b^2+1=2(2c^2+2c+1)$. Then $a^4=c(c+1)(2c^2+2c+1)$. $\gcd(c,c+1)=1$ $\gcd(c, 2c^2+2c+1)=\gcd(c,(c+1)^2+c)=\gcd(c,(c+1)^2)=1$ $\gcd(c+1, 2c^2+2c+1)=\gcd(c+1,(c+1)^2+c)=\gcd(c+1,c)=1$ Hence all of $c$, $c+1$ and $2c^2+2c+1$ are fourth powers, i.e. $c$ and $c+1$ are consecutive squares. This implies $c=0\implies a=0 \implies z=0 \implies y=1$. This gives the solution $(0,1)$. Case 2 $z=a^4$ and $z+1=8b^4$ Then $a^4+1=8b^4$. Looking modulo $5$, the LHS is one of $1,2$ while the RHS is one of $0,3$, a contradiction. Thus the only solution is $(0,1)$, however since we assumed $y$ is positive, there is also the solution $(0,-1)$.

How can one miss this solution? Consider the pell's equation $y^2-2z^2=1$ . We need to find when $z$ is a perfect square. Note that $(y,z)=(3,2)$ is a solution and we have all solutions from this. $Y+\sqrt2 Z=(3+2\sqrt2)^N=(1+\sqrt2)^{2N}$ this gives $Z= 2^{N-1}+2^{N-2}+....+2+1=2^N-1$ and $2^N-1$ is a perfect square only when Z=0,1 so the solution are $(0,1) (0-1)$