Difference between revisions of "2023 IOQM/Problem 4"
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<cmath>x^{4}=(x-1)(y^{3}-23)-1</cmath> | <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> |
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x⁴=(x-1)(y³-23)-1 | x⁴=(x-1)(y³-23)-1 | ||
x⁴-1=(x-1)(y³-23)-2 | x⁴-1=(x-1)(y³-23)-2 | ||
Revision as of 23:30, 31 August 2024
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
