2020/080) Solve in integer x $5^x - 3^x = 16$

Working mod 3 we have $(-1)^x\equiv 1 \pmod 3$

So x is even say 2n

So we have $^{2n} - 3^{2n} = 16$

Or $(5^n+3^n)(5^n-3^n) =16$

On  LHS both are even and unequal and 1st term is larget and 16 need to be factored into 2 even numbers 8 * 2

So $5^n +3^n = 8$

$5^n - 3^n = 2$

Adding we get $2* 5^n = 10$ or n = 1 so x = 2