Difference between revisions of "2023 IOQM/Problem 4"

Revision as of 23:29, 31 August 2024

Problem

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

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

Page

Toolbox

Search

Difference between revisions of "2023 IOQM/Problem 4"

Revision as of 23:29, 31 August 2024

Problem

@@ Line 4: / Line 4: @@
 Find the maximum possible value of <math>x + y</math>.
-==Solution1(Diophantine)==
+x⁴=(x-1)(y³-23)-1
-<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
+x⁴-1=(x-1)(y³-23)-2
+(x²-1)(x²+1)=(x-1)(y³-23)-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
+(x-1)(x+1)(x²+1)=(x-1)(y³-23)-2
+(x+1)(x²+1)=(y³-23)-(2⁄x-1)
-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.
+x≠1, x is an integer so x-1|2
+thus x-1≼2, x≼3, thus x= 2 or 3
-So, <math>x+y=3+4=\boxed{7}</math>
+For x=2 , (2+1)(4+1)=(y³-23)-(2/2-1)
+(3)(5)=(y³-23)-2
-~SANSGANKRSNGUPTA(inspired by PJ AND AM sir)
+=(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
++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