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Since $3^{x} +x^{4} =y!+2019 >2019$, we have $x\geq 6$.
Assume that $y\geq 6$, then $x^{4} \equiv 3\bmod 9$, impossible.
So we must have $y\leq 5$.
Exponential Balance
Find the sum of all positive intengers $x,y$,such like $[(x+y)]$ satisfying:
$$3^x+x^4=y!+2019.$$
Equation
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