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Difference between revisions of "2023 IOQM/Problem 4"

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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)==
 
==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
+
<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  
  
 +
<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 
  
−
<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 
+
This means <math>x=0</math>, <math>-1</math> (rejected as <math>x</math> is a [[positive integer]]), <math>2</math>, <math>3</math>. Thus, <math>x=2</math> or <math>3</math>. Now checking for each value, we find that when <math>x=2</math>, there is no [[integral]] value of <math>y</math>. When <math>x=3</math>, <math>y</math> evaluates to <math>4</math> which is the only possible positive [[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 positive [[integral]] solution.
+
So, <math>x+y=3+4=\boxed{7}</math>
−
 
 
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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)

Latest revision as of 07:53, 26 June 2024

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 $x-1$ (considering that $x\neq 1$) 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), $2$, $3$. Thus, $x=2$ 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 positive integral solution.

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

~SANSGANKRSNGUPTA(inspired by PJ AND AM sir)

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