Difference of Two Squares

Number Theory Level 2

S = { a 2 − b 2 ∣ a , b are positive integers } S = \{ a^2 - b^2 | a, b \text{ are positive integers} \}

How many values of S S are prime?

0 Infinitely many Finitely many

Chung Kevin by Chung Kevin , 46, Australia

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1 solution

Yatin Khanna
Oct 30, 2016

Despite the fact that a 2 − b 2 a^2 - b^2 can be factorised as ( a + b ) ( a − b ) (a+b)(a-b) ; it does not ensure that a 2 − b 2 a^2-b^2 is composite due to the fact that a − b a-b can take the value 1; and at the same time a + b a+b can take the value of a prime.

This can be easily achieved. For any odd prime p p ; we can form a pair of such variables.
a = p − 1 2 + 1 , b = p − 1 2 a = \frac {p-1}{2} + 1, b = \frac {p-1}{2}

And since there are infinite primes; there are infinite pairs.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

I wonder if others would see the factorization ( a − b ) ( a + b ) (a-b)(a+b) and think that the numbers will have to be composite.

Chung Kevin - 4 years, 7 months ago

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In fact, I did that too but I realized in time that (a-b) may even be 1 and hence; a^2 - b^2 may not be composite.

Yatin Khanna - 4 years, 7 months ago

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That's what led me to set this problem :)

Can you add that observation into the solution? It would help those who answered wrongly understand the mistake that they made.

Chung Kevin - 4 years, 7 months ago

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@Chung Kevin – IInteresting, i always start with looking at that identity. But when it comes to squares i get triggered by another fact. 2 following squares are an odd number apart. Since there are infinitely many odd primes, you can find infitely many pairs off following squares which are a prime apart.

Peter van der Linden - 4 years, 7 months ago

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@Peter van der Linden – Ah yes. Here's a followup.

Chung Kevin - 4 years, 7 months ago

@Peter van der Linden – You know what; I had added the same thing as solution number 2 but somehow I thought that both the solutions were quite similar so I never clicked on the publish button.

Yatin Khanna - 4 years, 7 months ago

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@Yatin Khanna – Well yours is more elegant and gives a way to find those pairs :)

Peter van der Linden - 4 years, 7 months ago

I've edited the solution slightly. I think stressing "despite the fact" is slightly more important.

Also, note that the pairing only works for odd primes.

Calvin Lin Staff - 4 years, 7 months ago

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Thank you Sir!

Yatin Khanna - 4 years, 7 months ago

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