Question

Is this very easy?

Given that p, q are prime numbers such that \(p^2+q = 37q^2+p\). find \( p−q\) ?

Answer 1

This problem is added only on the request of Mr.Math.

Answer 2

p=43 and q=7
p-q=43-7=36

Answer 3

The difference between two primes (none of which is 2) is an even number. Someone can easily see that 2 can't be a solution for either p nor q. So let \(p-q=2k, k\in \mathbb Z\), then:
\[p^2-37q^2=2k \implies 2k(p+q)-36q^2=2k\]
\[ \implies 2k=\frac{36q^2}{p+q-1} \implies k=\frac{18q^2}{p+q-1}.\]
Sice k is an integer then \((p+q-1)|18q^2 \implies (p+q-1)|18 \text{ or }(p+q-1)|q^2.\)

For (p+q-1)|18, there are few cases here in which none is correct. If (p+q-1)|q^2, then p+q-1 is equal 1, q or q^2. It's very obvious that the only possible case is that \(p+q-1=q^2\). Thus \(k=\frac{18q^2}{q^2}=18 \implies 2k=p-q=36\).

Answer 4

If you are assuming \( p-q=2k, k\in \mathbb Z \), how can you say that \( p^2-37q^2=2k \) same 2k ?

Answer 5

\[p^2+q=37q^2+p \implies p^2-37q^2=p-q=\cdots\]

Answer 6

Yes good, you should include every step, you know that I am a FOOL :P

Answer 7

Not bad Mr.Math ;)

Answer 8

Thanks Mr.Fool ;D

Answer 9

Neat solution Mr.Math

Answer 10

Thanks asnaseer! :-)

Answer 11

I guess I'm learning from you :)

Answer 12

:-) - I doubt that Sir! You are far cleverer than you think.