Number Theory on Christmas

6 a = p 2 + 1 6^a=p^2+1 where a a is a positive integer and p p is a prime.

Find the number of ordered pairs of ( a , p ) (a,p) which satisfy the above expression.

0 2 6 Infinitely many solutions 3 1

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2 solutions

Kushal Bose
Dec 21, 2016

For a prime p p it may be written as p = 6 m ± 1 p=6m \pm 1

So p 2 1 ( m o d 6 ) p^2 \equiv 1 \pmod{6} .Left hand side is always divisible by 6 6 .But in right hand side remainder will be always 2 2 .So there will be no solution

You may want to specify that p p must be a prime integer if that is what you intended. However, the answer is in fact still 0 0 even if p p can be any positive integer, (although the proof is just a bit more complicated).

Brian Charlesworth - 4 years, 5 months ago

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If p is any integer then p 0 , 1 , 2 ( m o d 3 ) p 2 0 , 1 ( m o d 3 ) p 2 + 1 1 , 2 ( m o d 3 ) p \equiv 0,1,2 \pmod{3} \\ p^2 \equiv 0,1 \pmod{3} \\ p^2+1 \equiv 1,2 \pmod{3} but

6 a 0 ( m o d 3 ) 6^a \equiv 0 \pmod{3} so there is no solution

Kushal Bose - 4 years, 5 months ago

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Nice approach!

Brian Charlesworth - 4 years, 5 months ago

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@Brian Charlesworth What is ur method ?

Kushal Bose - 4 years, 5 months ago

Yes I have updated

Kushal Bose - 4 years, 5 months ago
Spandan Senapati
Feb 14, 2017

Any a we choose the left side always factors out.but then this must be the product of two same primes.this means that both the factors must be equal.this is clearly a contradiction.

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