<s>[INST] Find all the $a$ such $539|a3^{253}+5^{44}$ This is what i thought:
Given that  $539|a3^{253}+5^{44}$ then $11|a3^{253}+5^{44}$ and $7^2|a3^{253}+5^{44}$
using congruences I get:
$$a3^{253}+5^{44} \equiv 0 \pmod{7^2}$$
and 
$$a3^{253}+5^{44} \equiv 0 \pmod{11}$$
then as a condition, I get $(a , 11)\not=1$ and $(a, 49)\not= 49$
given that $(3 , 7^2) = 1$ and $(5 , 7^2) = 1$ 
$$a3+5^{2} \equiv 0 \pmod{7^2}$$
$$a3 \equiv -5^{2} \equiv 24 \pmod{7^2}$$
$$a \equiv 8 \pmod{7^2}$$
Doing the same for $11$
$$a3^{253}+5^{44} \equiv 0 \pmod{11}$$
$$a3^{3}+5^{4} \equiv 0 \pmod{11}$$
$$a3^{3} \equiv 2 \pmod{11}$$
$$a \equiv 7 \pmod{11}$$
then whith this two conditions, using the Chinese remainder theorem:
$$a \equiv 502 \pmod{539}$$
Is this correct? what happens if $(a, 49) = 7$?
 [/INST] $539=7^2\cdot11$. Per Fermat's Little Theorem, we have $3^{10}\equiv1\text{ mod }11\iff3^{253}\equiv3^3\equiv5\text{ mod }11$ and $5^{10}\equiv1\text{ mod }11\iff5^{44}\equiv5^4\equiv9\equiv-2\text{ mod }11\iff5a-2\equiv0\text{ mod }11\iff5a\equiv2$ $\text{mod }11\iff a\equiv7\text{ mod }11$: The first condition.

By brute force, $5^{21}\equiv-1\text{ mod }49\iff5^{44}\equiv5^2=25\text{ mod }49$, and $3^{42}\equiv1\text{ mod }49\iff3^{253}$ $\equiv3^1=3\text{ mod }49\iff3a+25\equiv0\text{ mod }49\iff3a\equiv24\text{ mod }49\iff a\equiv8\text{ mod }49$: The second condition.

$11x+7=49y+8\iff11x=49y+1\iff11(x-4y)=5y+1\iff y=2$ and $x=9$, thus $a\equiv106\text{ mod }539$. Indeed, the computer can confirm the result.
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