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| | 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
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| − | x⁴-1=(x-1)(y³-23)-2
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| − | (x²-1)(x²+1)=(x-1)(y³-23)-2
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| − | (x-1)(x+1)(x²+1)=(x-1)(y³-23)-2
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| − | (x+1)(x²+1)=(y³-23)-(2⁄x-1)
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| − | x≠1, x is an integer so x-1|2
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| − | thus x-1≼2, x≼3, thus x= 2 or 3
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| − | For x=2 , (2+1)(4+1)=(y³-23)-(2/2-1)
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| − | (3)(5)=(y³-23)-2
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| − | 15=(y³-23)-2
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| − | y³= 15+2+23
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| − | y³=40, but y is an integer and 40 is not an perfect cube
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| − | thus x≠2
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| − | For x=3 , (3+1)(9+1)=(y³-23) - (2/3-1)
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| − | (4)(10)=(y³-23)-1
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| − | 40+1=y³-23
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| − | y³=41+23
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| − | y³=64, y=4
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| − | thus , x=3,y=4 , so x+y= 3+4=7
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| − | So the answer of this question will be 7
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Latest revision as of 23:30, 31 August 2024
be ![\[x^{4}=(x-1)(y^{3}-23)-1\]](http://latex.artofproblemsolving.com/5/0/9/509f87eace1256402c53d3e8ceac7a97e2c01ef0.png)
