2023 IOQM/Problem 4

Problem

Let $x, y$ be positive integers such that $x^{4}=(x-1)(y^{3}-23)-1$

Find the maximum possible value of $x + y$ .

Solution1(Diophantine)

$x^{4}=(x-1)(y^{3}-23)-1$ , subtracting 1 on both sides we get $x^{4}- 1^{4}=(x-1)(y^{3}-23)-2$ . Factorizing the LHS we get

$(x^{2}+ 1)(x-1)(x+1) =(x-1)(y^{3}-23)-2$ . Now divide the equation by $x-1$ (considering that $x\neq 1$ ) to get $(x^{2}+ 1)(x+1) = (y^{3}-23)-\frac{2}{x-1}$ Since $x$ and $y$ are integers, this implies $x-1$ divides 2, so possible values $x-1$ are -1, -2, 1, 2

This means $x=0$ , $-1$ (rejected as $x$ is a positive integer), $2$ , $3$ . Thus, $x=2$ or $3$ . Now checking for each value, we find that when $x=2$ , there is no integral value of $y$ . When $x=3$ , $y$ evaluates to $4$ which is the only possible positive integral solution.

So, $x+y=3+4=\boxed{7}$

~SANSGANKRSNGUPTA(inspired by PJ AND AM sir)

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2023 IOQM/Problem 4

Problem

Solution1(Diophantine)