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Number theory Difficulty 4.6 AIME Prove it Saudi Arabia

Prove that there are no positive integers xx, yy, zz such that
(3x+4y)(4x+5y)=7z (3x + 4y)(4x + 5y) = 7^{z}

Solution

From the condition, we can see that both numbers 3x+4y3x + 4y and 4x+5y4x + 5y are powers of 77, so such must be also their division. However
1<4x+5y3x+4y<2 1 < \frac{4x + 5y}{3x + 4y} < 2
and can't be power of 77.

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