Solve the equation $3^{x}+4^{y}=5^{z}$ in natural numbers.

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Consider the remainders of division by 3 and 4, and use the difference of squares formula.

## Solution

The right-hand side, when divided by 3, should give the same remainder as the left-hand side, which is 1. Therefore, \( z \) is even.

Similarly, the left-hand side, when divided by 4, gives a remainder of 1, so \( x \) is also even. Thus, \( 4^{y} = 5^{z} - 3^{x} = 5^{2u} - 3^{2v} \), which means \( 2^{2y} = (5^{u} - 3^{v})(5^{u} + 3^{v}) \). Both terms on the right are powers of 2. Let \( 5^{u} - 3^{v} = 2^{k} \) and

\( 5^{u} + 3^{v} = 2^{l} \), where \( k, l \geq 0 \) and \( k + l = 2y \). Then \( 5^{u} = \frac{1}{2}(2^{k} + 2^{l}) \) and \( 3^{v} = \frac{1}{2}(2^{l} - 2^{k}) \). Therefore, \( l > k \), so \( 2^{l} \) is divisible by 4, and \( 2^{k} \) is not, meaning \( k = 1 \), \( 2^{k} = 2 \),

and \( 3^{v} = 2^{l-1} - 1 \). This implies that \( l-1 \) is even, \( l-1 = 2s \). Then \( 3^{v} = (2^{s} - 1)(2^{s} + 1) \) — the product of two numbers differing by 2 and being powers of 3. Therefore, these factors are 1 and 3. Hence, \( s = 1 \), \( l = 3 \), \( 2y = 4 \).

## Answer

\((2, 2, 2)\)