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Problem 1031

AMC 12 late, AIME early

Number theory Difficulty 4.9 Prove it Belarusian Mathematical Olympiad · Belarus

Find all triples of positive integers $(x, y, z)$ satisfying the equality $3^x + 7^y = 4^z$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

$z \ge 3$ , then $2^z - 1 \vdash 3$ , but it is possible only if $z$ is even. Let $z = 2c$ for some nonnegative integer $c$ . Then $4^c - 3^a = 1$ . If $a = 1$ , we find $c = 1$ and obtain $x = 1, z = 2, y = 1$ . If $a > 1$ , then $4^c = 3^a + 1$ has remainder 1 when divided by 9. The power of 4 has this remainder only if $c$ is divisible by 3. But in this case $4^3 \equiv 1 \pmod 7$ and hence $4^c \equiv 1 \pmod 7$ . It follows that $3^a = 4^c - 1 \vdash 7$ , which is impossible.

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