There are no positive integers such that $a+b^2$, $a^2+b$ are both powers of $5$ Prove that there doesn't exist positive integers $a$ and $b$ such that $a^2+b$ and $b^2+a$ are both powers of $5$.
Assume on the contrary. Let $$a^2+b=5^x, \quad a+b^2=5^y$$
With $x,y\ge 1$. WLOG assume $x\ge y$ which means $a\ge b$ and $$a+b^2\mid a^2+b\implies a^2+b\equiv 0\pmod{a+b^2}$$
Since $a\equiv -b^2\pmod {a+b^2}$ we get $$a^2+b\equiv b^4+b\equiv 0\pmod{a+b^2}$$
Claim $(a,5)=(b,5)=1$. Furthermore $(a,b)=1$.
Proof
Assume $5\mid a$ then immediately $5\mid b$. Let $a=5a_1,$ $ b=5b_1$ plugging this to the original equation $$5a_1^2+b_1=5^{x-1}$$
Now if $x=1$ we get a contradiction. So assume $x\ge 2$ clearly $5\mid b_1$ meaning $b_1=5b_2$. $$a_1^2+b_2=5^{x-2}$$
Again $x=2$ is impossible. So $a_1=5a_2...$ Cleary this can't go on forever hence $a_1=a_2=0$ a contradiction.
If $p\mid (a,b)$ then $p\mid 5$ meaning $p=5$ contradiction.
Back to our result $a+b^2\mid b(b^3+1)$ since $(a+b^2,b)=1$ we get $$a+b^2\mid b^3+1$$
We're pretty much done. we just need the following claim.
Claim $5\mid b+1$.
Proof We know $5\mid b^3+1$ Thus $$b^3\equiv -1\pmod 5\implies b^6\equiv 1\pmod 5$$
$\operatorname{ord}_5b\mid 6$ but $\operatorname{ord}_5b \mid 4$ so $\operatorname{ord}_5b=2$ or $1$. Surely it's not $1$.
$$b^2\equiv 1\pmod 5 \implies  b\equiv -1 \pmod 5$$  So $5\mid b+1$.
All our work till now was in order to use the LTE lemma (Lifting the exponent). Since $5\nmid b$ and $5\mid b+1$ we get $$\nu_5(a+b^2)\le \nu_5(b^3+1)=\nu_5(b+1)+\nu_5(3)=\nu_5(b+1)$$
In fact since the LHS is a power of $5$ we get $a+b^2\mid b+1 \implies b+b^2\le a+b^2\le b+1$. Hence $b\le 0$ A contradiction.
I just need a verification and if you see a part where I could've used a better method to do it -especially the first claim- please tell me.

Let be by contradiction $a^2+b=5^x$ and $a+b^2=5^y$ with $x\lt y$ (one has $x\ne y$ necessarily) so we have $a^2+b\equiv0\pmod{5^k}$ for $k=1,2,\cdots, x$ and similarly $a+b^2\equiv0\pmod{5^k}$.
For $k=1$ we have the table of the function $f(x,y)=x^2+y$ modulo $5$ where one can see that the only $(a,b)$ and $(b,a)$ giving zero is $(4,4)$. Explaining we have $(x,y)=(1,4),(2,1),(3,1)$ makes $0$ modulo $5$ but $(4,1),(1,2),(1,3)$ does not.
$$\begin{array}{|c|c|}\hline &1 & 2 & 3 & 4 &0 \\\hline 1 & 2 & 3 &4&0&1\\\hline 2 &0&1&2&3&4 \\\hline 3&0&1&2&3&4\\\hline4&2&3&4&0&1\\\hline0&1&2&3&4&0\\\hline\end{array}$$
It follows that $a=5A-1$ and $b=5B-1$ (because $4=-1\pmod5$).
Come back to the beginning we have $$\begin{cases}a^2+b=5^x\\a+b^2=5^y\end{cases}\Rightarrow\begin{cases}5^2A^2-10A+5B=5^x\\5^2B^2-10B+5A=5^y\end{cases}\Rightarrow\begin{cases}5A^2-2A+B=5^{x-1}\\5B^2-2B+A=5^{y-1}\end{cases}$$
from which $$(B-A)\left[5(B+A)-3\right]=5^{y-1}-5^{x-1}=5^{x-1}(5^{y-x}-1)$$ so we have $$B-A=5^{x-1}\\5(B+A)=5^{y-x}+2\hspace1cm(*)$$ The second equation $(*)$ leads to the absurd $2\equiv0\pmod5$ or $3\equiv0\pmod5$. This finishes the proof.
