Question

Prove that there are no solutions in positive integers\(x\)and\(y\)to the equation\({x^4} + {y^4} = 625\).

Short answer

There are no solutions of the equation\({x^4} + {y^4} = 625\)for positive values x and y.

Step-by-step solution

Introduction

No solution occurs for a given equation if no value of the variables in the equation is satisfied for the given condition.

Proof for the given equation to be false

The given equation is \({x^4} + {y^4} = 625\).

It can be written as:\({x^4} + {y^4} = {5^4}\)

If the solutions of these equations have to be positive integers, then the values must be less than 5.

As the fourth power of the integer is not of negative value, it is required to check the positive integers having fourth power as smaller than 625.

The fourth power of 1, 2, 3, 4, 5 are:1, 16, 81, 256, 625.

Thus, each of the terms can only have maximum value, \({4^4} = 256\).

Checking various possible combination of fourth power

\(\begin{aligned}{}1 + 1 = 2\\ \ne 625\end{aligned}\)

\(\begin{aligned}{}1 + 16 = 17\\ \ne 625\end{aligned}\)

\(\begin{aligned}{}1 + 81 = 82\\ \ne 625\end{aligned}\)

\(\begin{aligned}{}1 + 256 = 257\\ \ne 625\end{aligned}\)

\(\begin{aligned}{}16 + 1 = 17\\ \ne 625\end{aligned}\)

\(\begin{aligned}{}16 + 16 = 32\\ \ne 625\end{aligned}\)

\(\begin{aligned}{}16 + 81 = 97\\ \ne 625\end{aligned}\)

\(\begin{aligned}{}16 + 256 = 272\\ \ne 625\end{aligned}\)

\(\begin{aligned}{}81 + 1 = 82\\ \ne 625\end{aligned}\)

\(\begin{aligned}{}81 + 16 = 97\\ \ne 625\end{aligned}\)

\(\begin{aligned}{}81 + 256 = 337\\ \ne 625\end{aligned}\)

\(\begin{aligned}{}256 + 1 = 257\\ \ne 625\end{aligned}\)

\(\begin{aligned}{}256 + 16 = 272\\ \ne 625\end{aligned}\)

\(\begin{aligned}{}256 + 81 = 337\\ \ne 625\end{aligned}\)

\(\begin{aligned}{}256 + 256 = 512\\ \ne 625\end{aligned}\)

This condition is such that the addition does not exceed 625 as in the above case, the addition will be only 512.

Thus, the required result is found.