Not Catalan's Conjecture

Number Theory Level 3

x 5 + 31 = y 2 \large x^{5} + 31 = y^{2}

How many integer solutions does the equation above has?


The answer is 0.

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Danish Ahmed by Danish Ahmed , 24, India

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3 solutions

Discussions for this problem are now closed

Curtis Clement
Mar 22, 2015

y 2 ≡ 1 , 3 , 4 , 5 o r 9 ( m o d 11 ) y^2 \equiv 1,3,4,5 \ or \ 9 (mod11) x 5 ≡ 1 , 10 ( m o d 11 ) ⇒ x 5 + 31 ≡ 8 , 10 ( m o d 11 ) x^5 \equiv 1 , 10 (mod 11) \Rightarrow\ x^5 +31 \equiv 8 \ , \ 10 (mod 11) ∴ b y c o n t r a d i c t i o n t h e r e a r e n o s o l u t i o n s \large\therefore\ \ by \ contradiction \ there \ are \ no \ solutions

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William Isoroku
Feb 23, 2015

Just graph it

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

This solution is marked wrong. Even if you graphed it, you would only conclude that there is no solution for a finite number of integers. When in fact we want to find ALL integer solution(s) to the diophantine equation.

How does graphing ensure that there can be no solutions for x , y x,y over the integers? Sure, you can verify within a limited range that there are no integer solutions but that doesn't guarantee the non-existence of solutions over the whole set of integers.

Prasun Biswas - 6 years, 3 months ago

consider factors of type 4k+1 and 4k+3

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

This solution has been marked wrong. Despite not showing any relevant work and taking into consideration of your comment below, you cannot show that there's no solution by simply factoring x 5 + 32 x^5 + 32 .

What are they supposed to be factors of? x 5 + 31 x^5 +31 ?

Sudeshna Pontula - 6 years, 3 months ago

Jajaja, wait a minute, add one to each side, then the RHS should have its 4k+3 primes elevated to even powers, etc you can factorize x^5+32... assume some form for x+2 work out the details and get contradictions :)

Héctor Andrés Parra Vega - 6 years, 3 months ago

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