1

Find all the positive integers p,q,n such that p,q are primes and also $$(p^2 + 1)(q^2 + 1) = n^2 + 1$$.

We know that pair squares are all divisible by 4 (because they have two or more times 2 as factor), so I squared the possible residues of 4. $$(4k)^2 = 16k^2 = 4k$$ $$(4k+1)^2 = 16k^2 + 8k + 1 = 4k+1$$ $$(4k+2)^2 = 16k^2 + 16k + 4 = 4k$$ $$(4k+3)^2 = 16k^2 + 24k + 9 = 4k+1$$

So, every squared odd number has residue 1 when it's divided by 4. We can notice that $n^2 + 1$ has residue 1 or 2. Let's suposse $p$ and $q$ are both odd numbers, then $ (p^2 + 1)(q^2 + 1) $ Has to be divisible by 4 (because both factors let residue 2), and that doesn't make sense. So, $p$ or $q$ (it doesn't matter) has to be pair, an the only pair prime is 2. Then we have $$5(q^2 +1) = n^2 + 1$$

I found $$5(3^2 + 1) = 7^2 + 1$$ but i don't know if they're the only 3 numbers. So i'm wondering to know how to prove that they're, and if they aren't wich are the others (and again, how to prove that they're the only) using the method i did or another (but not that complicated)

Thanks.

SonodaUmi
  • 660

1 Answers1

7

We have to solve $p^2+p^2 q^2+q^2 = n^2$. If both $p$ and $q$ are odd, the LHS is $\equiv 3\pmod{4}$, hence it cannot be a square. It follows that the only solutions are given by $p=2$ and $$ 5q^2+4=n^2 $$ whose solutions depend on Fibonacci numbers of the form $F_{2m}$. The only Fibonacci number of the form $F_{2m}$ that is also a prime is $F_4=3$ (due to $F_{2n}=F_n L_n$), hence the only solutions are $$ (p,q,n)\in\{(2,3,7),(3,2,7)\}.$$

Jack D'Aurizio
  • 361,689
  • How do you know the solutions are the Fibbonacci numbers? (i opened the pdf but i didn't get it, i'm not that good at number theory, can you please try to explain a little? – SonodaUmi Feb 26 '17 at 21:02
  • @Pulie: that is a well-known topic. The solution of Pell's equations $5n^2\pm 4=m^2$ are given by Fibonacci numers. That also follows from the general theory of Pell's equations. – Jack D'Aurizio Feb 26 '17 at 21:08
  • See also https://en.wikipedia.org/wiki/Fibonacci_number#Identification – lhf Dec 02 '21 at 20:36