If $y$ is negative, $LHS$ is an integers whereas $RHS$ isn't, which is impossible. If $y = 0$, we get $x = 1$. If $y > 0$, then using Fermat's Little Theorem we get $101 \mid x^{100} - (-2)^{20}$ i.e. $101 \mid 1 - 1024^2$ meaning $101 \mid 1 - 14^2 = 1 - 196$ from where we get $101 \mid 7$, which clearly doesn't hold, implying the only solution is $(x, y) \in$ {$(1, 0)$}

If $y>0$ then $x^5\equiv -2\text{(mod 101)}\implies x^{100}\equiv 95\text{(mod 101)}$, a contradiction. Thus the only solution is $\boxed{(x,y)=(1,0)}$

kootrapali wrote: If $y>0$ then $x^5\equiv -2\text{(mod 101)}\implies x^{100}\equiv 95\text{(mod 101)}$, a contradiction. Thus the only solution is $\boxed{(x,y)=(1,0)}$ why is $x^{100}\equiv 95\text{(mod 101)}$ a contradiction?

adambd wrote: kootrapali wrote: If $y>0$ then $x^5\equiv -2\text{(mod 101)}\implies x^{100}\equiv 95\text{(mod 101)}$, a contradiction. Thus the only solution is $\boxed{(x,y)=(1,0)}$ why is $x^{100}\equiv 95\text{(mod 101)}$ a contradiction? Because by Fermat's little theorem, we have $x^{100} \equiv 1 \pmod{101}$.

Lukaluce wrote: Solve the following equation in the set of integers $x^5 + 2 = 3 \cdot 101^y$. Clearly $y<0$ has no solutions since an integer well be equal to a pure rational. So $y \ge 0$. Assume that $y>0$ and then use $\pmod{101}$ $$x^5 \equiv -2 \pmod{101} \implies x^{100} \equiv (-2)^{20} \pmod{101}$$Now we will se what is $2^{20}$ on $\pmod{101}$ $$2^{20}=2^7 \cdot 2^7 \cdot 64 \equiv 27^2 \cdot 64 \equiv 22 \cdot 64 \equiv 2 \cdot -3 \equiv 95 \pmod{101}$$By theorem of euler totient function we have: $$x^{\phi(101)} \equiv 1 \pmod{101} \implies x^{100} \equiv 1 \pmod {101}$$Both equations mean: $$1 \equiv 95 \pmod{101} \; \text{contradiction!!}$$Then $y=0$ and plug it on the equation to get: $$x^5+2=3 \cdot 101^0 \implies x^5=1 \implies x=1$$Then the only pair is $(1,0)$ Thus we are done

Lukaluce wrote: Solve the following equation in the set of integers $x^5 + 2 = 3 \cdot 101^y$. OlympusHero wrote:

my solution is similar, practically the same

Yes, this is the solution most people thought of. Kootrapali also had this solution.

a little bit different

If $y$ is positive then $x^{100} \equiv 95 \pmod{101}$, but this is not possible because by FLT it needs to be 1 mod 101, so the only solution is $\boxed{(1,0)}$