Today my friend asked a question for help:
Find all solutions of $2^x=5^y+3$ for non-negative integers $x,y$.
It is obvious that the solutions are $(x,y)=(2,0),(3,1),(7,3)$, and I think there is no more solution. However, we can't prove that these are the only solutions. We have already tried to mod many numbers and still get "may" possible solutions other than the $3$ solutions I have written. I have surfed the Net and still can't find solutions. I hope you guys can help my friend solve. Thank you very much!
We start by changing the letters, plugging in your maximal exponents, and arriving at $$ 128 (2^x - 1) = 125 (5^y-1) \; . \; $$ We ASSUME that both $x,y \geq 1$ and get a contradiction. The main tasks are factoring $p^n - 1$ and finding the order of a prime for some target $n,$ the minimal $k$ such that $p^k \equiv 1 \pmod k$
So, order_2 of 125 is 100, where we have simply $\varphi(125) = 100.$ Thus $100|x.$
Next $$ 2^{100} - 1 = 3 \cdot 5^3 \cdot 11 \cdot 31 \cdot 41 \cdot 101 \cdot 251 \cdot 601 \cdot 1801 \cdot 4051 \cdot 8101 \cdot 268501 $$ We need $5^y \equiv 1 \pmod {268501}$ Therefore $125 | y.$ $$ 5^{125} - 1 = 2^2 \cdot 11 \cdot 71 \cdot 101 \cdot 251 \cdot 401 \cdot 3597751 \cdot 9384251 \cdot \mbox{BIG} $$ We need $2^x \equiv 1 \pmod { 9384251}$ Therefore $125 | x.$ From the beginning, we have had $100 | x.$ Put them together, Chinese Remainder Theorem, and we arrive at $$ 500 | x $$
And $$ 2^{500}-1 = (2^{100}-1)(2^{400}+2^{300}+2^{200}+2^{100}+1)$$ where $2^{100}-1$ is a multiple of $5^3$ (above) and the second factor is a multiple of $5$ because $2^{100}=16^{25}\equiv 1\bmod 5$. Thus $2^{500} - 1$ is a multiple of $5^4.$ We have gotten what we wanted, as
$ 128 (2^x - 1) = 125 (5^y-1) \; \; $ tells us that $ 125 (5^y-1)$ is divisible by $625.$ In turn, this says that $ (5^y-1)$ is divisible by $5.$ This is impossible with $y \geq 1,$ so that $y = 0,$ contradicting our hypothesis.