Difference between revisions of "2023 IOQM/Problem 4"

Latest revision as of 23:30, 31 August 2024

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

Find the maximum possible value of $x + y$

@@ Line 3: / Line 3: @@
 <cmath>x^{4}=(x-1)(y^{3}-23)-1</cmath>
-Find the maximum possible value of <math>x + y</math>.
+Find the maximum possible value of <math>x + y</math>
-==Solution1(Diophantine)==
-<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
-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.
-So, <math>x+y=3+4=\boxed{7}</math>
-~SANSGANKRSNGUPTA(inspired by PJ AND AM sir)