20 + 124 = 52
100 + 528 = 232
=> x=0, y=2
\(a,5^y-2^x=124\)
Do \(124\equiv-1\left(mod5\right)\)
Mà \(5^y\equiv0\left(mod5\right)\)
\(\Rightarrow2^x\equiv-1\left(mod5\right)\)
Vậy x chẵn
Với x = 2k
\(\Rightarrow5^y-2^{2k}=124\)
có \(124\equiv0\left(mod4\right)\)
\(4^k\equiv0\left(mod4\right)\)
\(\Rightarrow5^y\equiv0\left(mod4\right)\)
\(\Rightarrow y\)lẻ
\(\Rightarrow5^{2k+1}-2^{2k}=124\)
\(\Rightarrow5^{2k}-2^{2k}+5^{2k}.2^2=124\)
\(\Rightarrow\left(5^k-2^k\right)\left(5^k+2^k\right)+\left(5^k.2\right)^2=124\)