If $(a^2−b^2)$ is a prime number where $a$ and $b\in N$, then:

Question

$a^2−b^2 = 3$
$a^2−b^2 = a−b$
$a^2−b^2 = a+b$
$a^2−b^2 = 5$

My attempt:

$(a^2-b^2)=(a-b)(a+b)$

if $a + b = 1$ and $a^2 + b^2$ is prime. (Since $\cfrac{a^4 − b^4}{a- b} = (a + b)(a^2 + b^2)$.

Thus, in this case the pair $(a, b)$ must be $(x + 1, −x)$ and $x^2 + (x + 1)^2$ must be prime. That is, $x$ must be in OEIS A027861.)

I didn't find proper proof.

Can you explain it, please?

@Starfall, I think you can feel, if you voted down without proper reason? http://math.stackexchange.com/questions/1899159/is-the-difference-of-two-irrationals-which-are-each-contained-under-a-single-squ/1899216#comment3896542_1899216 — Mithlesh Upadhyay, Sep 15 '16 at 11:10

score 1 · Accepted Answer · answered Sep 15 '16 at 10:50

1

If $a^2-b^2$ is prime, either $a=1-b$ or $a=b+1$. Note that $a-b < a+b$, so $a+b = 1 \implies a-b<1$ is a contradiction, hence $a-b=1$, hence $a^2-b^2 = a+b$.

answered Sep 15 '16 at 10:50

Sarvesh Ravichandran Iyer

76,105

Thanks for nice explanation. – Mithlesh Upadhyay Sep 15 '16 at 10:57
1

@MithleshUpadhyay You are welcome. – Sarvesh Ravichandran Iyer Sep 15 '16 at 10:58

score 1 · Answer 2 · answered Sep 15 '16 at 10:53

1

If $a^2-b^2$ is a prime number, then $a-b$ must be $1$, because for any $a,b\in N$, $a+b\ne 1$. This directly leads to the result that $a^2-b^2=a+b$.

answered Sep 15 '16 at 10:53

codetalker

2,419

Thanks for nice explanation. – Mithlesh Upadhyay Sep 15 '16 at 10:56
1

Happy to help:) – codetalker Sep 15 '16 at 12:39

If $(a^2−b^2)$ is a prime number where $a$ and $b\in N$, then:

2 Answers2