Hint: $({{x}^{2}}-2{{y}^{2}})({{x}^{2}}+2{{y}^{2}})={{43}^{n}}\cdot {{47}^{n}}$, casework

Solve also this: Find all positive integers $n$ such that $1763^n$ can be expressed in the form $x^4+4y^4$ for some integers $x,y$. Edit:I can not find a solution with $2021$ but i find a solution for $1763$.

P2nisic wrote: Solve also this: Find all positive integers $n$ such that $2021^n$ can be expressed in the form $x^4+4y^4$ for some integers $x,y$. Sophie Germain Identity, and casework maybe(?). I havent done it so idk

Obviously, if $x=0$ then no solution and if $y=0$ then $n\equiv 0\pmod{4}$ is the only solution. Now, suppose that $x,y\neq 0$. Let $\gcd(x,y)=d$ and $x=da,y=db$. We have $$d^4(a^2-2b^2)(a^2+2b^2)=43^n\cdot 47^n.$$Let $d=43^k\cdot 47^l$ where $0\leq k,l\leq \frac{n}{4}$. Then $(a^2-2b^2)(a^2+2b^2)=43^{n-4k}\cdot 47^{n-4l}$. Claim 1: $\gcd(a^2-2b^2,a^2+2b^2)=1$ Proof. Assume the contrary that there is prime $p$ such that $p\mid a^2-2b^2,a^2+2b^2$. $$p\mid (a^2-2b^2)+(a^2+2b^2)\implies p\mid 2a^2$$If $p=2$ then $2\mid a^2-2b^2$ so $2\mid a$ which is absurd. Hence, $p\mid a$. $$p\mid (a^2+2b^2)-a^2\implies p\mid 2b^2\implies p\mid b$$Thus, $\gcd(a,b)\geq p$ which is a contradiction and the claim is proved. Claim 2: $47\nmid a^2+2b^2$ Proof. Assume the contrary, since $\gcd(a,b)=1$, we get that $\gcd(a,47)=\gcd(b,47)=1$. \begin{align*} a^2+2b^2\equiv 0\pmod{47} &\implies \bigg(\frac{a}{b}\bigg)^2\equiv -2\pmod{47} \\ &\implies \left(\frac{-2}{47}\right)=1 \\ &\implies (-1)^{\frac{47-1}{2}}\cdot (-1)^{\frac{47^2-1}{8}}=1 \\ &\implies -1=1 \end{align*}A contradiction. Thus, the claim is proved. After combining these claims, we have $a^2-2b^2=47^{n-4l}, a^2+2b^2=43^{n-4k}$. Since $a^2+2b^2>a^2-2b^2$, we get that $n-4k>0$. Thus, we have $$4b^2=(a^2+2b^2)-(a^2-2b^2)=43^{n-4k}-47^{n-4l}.$$So $43^r-47^s$ is a perfect square for some integer $r>0,s\geq 0$. Taking modulo $43$ gives us that \begin{align*} \left(\frac{43^r-47^s}{43}\right)=1 &\implies \left(\frac{-4^s}{43}\right)=1 \\ &\implies \left(\frac{-1}{43}\right)\left(\frac{4^s}{43}\right)=1 \\ &\implies (-1)^{\frac{43-1}{2}}\cdot \left(\frac{(2^s)^2}{43}\right)=1 \\ &\implies (-1)\cdot 1 = 1 \\ \end{align*}A contradiction. Hence, the only solution is $n=4m$ where $m$ is any positive integer.