Find four solutions to $p^2+1=q^2+r^2$ with primes $p$, $q$ and $r$?

Question

Q. Find four solutions to $p^2+1=q^2+r^2$ with primes $p$, $q$ and $r$?

My thinking: I think 2(= q or r ) will not satisfy this equation. Hence we are left with odd prime numbers. We can factories to to get (p+q-r-1)(p+r-q-1)=2(qr-p) Or simply (p+q)(p-q)=(r+1)(r-1) But I am not able to make any conclusions. It's is from {250 problems in elementary number theory} Q.78 (4.primes and composite numbers) Please give any hint(theorem that may help)

Answer 1

The question appears to allow $q=r$ and this then requires prime number solutions of Pell's equation $$p^2+1=2q^2$$

For example

$p=7, q=r=5$

$p=41, q=r=29$

Answer 2

To find all solutions systematically.

All solutions can be generated using primitive Pythagorean triples. Let $(a,b,c)$ be a primitive Pythagorean triple with $b$ even. Then $p,q,r$ form a solution if they are primes such that $$b=ap-cr,$$ $$bq=cp-ar.$$

Example (using the most obvious triple)

We must solve $$4=3p-5r,$$ $$4q=5p-3r.$$

Then there is a positive integer $k$ such that $$p=10k+3,q=8k+3,r=6k+1.$$

The relative abundance of these primes can be seen in the screenshot referenced below. (And this is for just one of infinitely many Pythagorean triples.)

[1]: https://i.stack.imgur.com/tOyfD.png

Answer 3

A nice identity could be $$\left(\left(\dfrac{(r-1)}{2}\right)^2+\dfrac{(r-1)}{2}+1\right)^2+1=\left(\left(\dfrac{(r-1)}{2}\right)^2+\dfrac{(r-1)}{2}-1\right)^2+r^2$$

Answer 4

There can be a solution when $(p^2+1)/2$ is composite. A way to find it by example:

$$13^2+1=170=17\cdot10=(4+i)(4-i)(3+i)(3-i)=\begin{cases}(13+i)(13-i)=13^2+1^2\text{ You know this solution}\\(11+7i)(11-7i)=11^2+7^2\text{ This is the new solution}\end{cases}$$

Alas, not always the other solution is with prime numbers