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| | <cmath>x^{4}=(x-1)(y^{3}-23)-1</cmath> | | <cmath>x^{4}=(x-1)(y^{3}-23)-1</cmath> |
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| − | Find the maximum possible value of <math>x + y</math>. | + | Find the maximum possible value of <math>x + y</math> |
| − | ==Solution1(Diophantine)==
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| − | <math>x^{4}=(x-1)(y^{3}-23)-1</math>, subtracting 1 on both sides we get <math>x^{4}- 1^{4}=(x-1)(y^{3}-23)-2</math> [[factorizing]] the LHS we get
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| − | <math>(x^{2}+ 1)(x-1)(x+1) =(x-1)(y^{3}-23)-2</math>. Now [[divide]] the [[equation]] by <math>x-1</math>(considering that <math>x\neq 1</math>) to get <cmath>(x^{2}+ 1)(x+1) = (y^{3}-23)-\frac{2}{x-1}</cmath> Since <math>x</math> and <math>y</math> are [[integers]], this implies <math>x-1</math> [[divides]] 2, so possible values <math>x-1</math> are -1, -2, 1, 2
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| − | This means <math>x</math>= 0, -1(Rejected as <math>x</math> is a [[positive integer]]), 2, 3. so <math>x</math> =2 or 3. Now checking for each value, we find that when <math>x</math>=2, there is no [[integral]] value of <math>y</math>. When <math>x</math>= 3, <math>y</math> evaluates to 4 which is the only possible [[integral]] solution.
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| − | So, <math>x+y</math>= 3+ 4 = <math>\boxed{7}</math>
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| − | ~SANSGANKRSNGUPTA(inspired by PJ AND AM sir)
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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)
