Difference between revisions of "2023 IOQM/Problem 4"

Latest revision as of 23:30, 31 August 2024

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

Find the maximum possible value of $x + y$

@@ Line 3: / Line 3: @@
 <cmath>x^{4}=(x-1)(y^{3}-23)-1</cmath>
-Find the maximum possible value of <math>x + y</math>.
+Find the maximum possible value of <math>x + y</math>
-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
-=(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
-+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