2023 IOQM/Problem 4
Problem
Let 
be positive integers such that
![\[x^{4}=(x-1)(y^{3}-23)-1\]](http://latex.artofproblemsolving.com/5/0/9/509f87eace1256402c53d3e8ceac7a97e2c01ef0.png)
Find the maximum possible value of 
.
x⁴=(x-1)(y³-23)-1
x⁴-1=(x-1)(y³-23)-2
(x²-1)(x²+1)=(x-1)(y³-23)-2
(x-1)(x+1)(x²+1)=(x-1)(y³-23)-2
(x+1)(x²+1)=(y³-23)-(2⁄x-1)
x≠1, x is an integer so x-1|2
thus x-1≼2, x≼3, thus x= 2 or 3
For x=2 , (2+1)(4+1)=(y³-23)-(2/2-1)
(3)(5)=(y³-23)-2
15=(y³-23)-2
y³= 15+2+23
y³=40, but y is an integer and 40 is not an perfect cube
thus x≠2
For x=3 , (3+1)(9+1)=(y³-23) - (2/3-1)
(4)(10)=(y³-23)-1
40+1=y³-23
y³=41+23
y³=64, y=4
thus , x=3,y=4 , so x+y= 3+4=7
So the answer of this question will be 7
