6 a = p 2 + 1 where a is a positive integer and p is a prime.
Find the number of ordered pairs of ( a , p ) which satisfy the above expression.
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You may want to specify that p must be a prime integer if that is what you intended. However, the answer is in fact still 0 even if p can be any positive integer, (although the proof is just a bit more complicated).
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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 ) but
6 a ≡ 0 ( m o d 3 ) so there is no solution
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Nice approach!
Yes I have updated
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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For a prime p it may be written as p = 6 m ± 1
So p 2 ≡ 1 ( m o d 6 ) .Left hand side is always divisible by 6 .But in right hand side remainder will be always 2 .So there will be no solution