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Difference between revisions of "2023 IOQM/Problem 4"

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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> 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   
 
<math>(x^{2}+ 1)(x-1)(x+1) =(x-1)(y^{3}-23)-2</math>.  Now [[divide]] the [[equation]] by <math>x-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   
  
−
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>=4  which is the only possible [[integral]] solution.  
+
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.  
  
 
So, <math>x+y</math>= 3+ 4 = <math>\boxed{7}</math>
 
So, <math>x+y</math>= 3+ 4 = <math>\boxed{7}</math>
  
 
~SANSGANKRSNGUPTA(inspired by PJ AND AM sir)
 
~SANSGANKRSNGUPTA(inspired by PJ AND AM sir)

Revision as of 10:17, 25 October 2023

Problem

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

Find the maximum possible value of $x + y$.

Solution1(Diophantine)

$x^{4}=(x-1)(y^{3}-23)-1$, subtracting 1 on both sides we get $x^{4}- 1^{4}=(x-1)(y^{3}-23)-2$ factorizing the LHS we get


$(x^{2}+ 1)(x-1)(x+1) =(x-1)(y^{3}-23)-2$. Now divide the equation by $x-1$ to get \[(x^{2}+ 1)(x+1) = (y^{3}-23)-\frac{2}{x-1}\] Since $x$ and $y$ are integers, this implies $x-1$ divides 2, so possible values $x-1$ are -1, -2, 1, 2

This means $x$= 0, -1(Rejected as $x$ is a positive integer), 2, 3. so $x$ =2 or 3. Now checking for each value, we find that when $x$=2, there is no integral value of $y$. When $x$= 3, $y$ evaluates to 4 which is the only possible integral solution.

So, $x+y$= 3+ 4 = $\boxed{7}$

~SANSGANKRSNGUPTA(inspired by PJ AND AM sir)

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