Find four solutions to $p^2+1=q^2+r^2$ with primes $p$, $q$ and $r$? 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)
 A: 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
$p=7, q=r=5$
$p=41, q=r=29$
A: 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
A: A nice identity could be $$\left(\left(\dfrac{(r-1)}{2}\right)^2+\dfrac{(r-1)}{2}+1\right)^2+1=\left(\left(\dfrac{(r-1)}{2}\right)^2+\dfrac{(r-1)}{2}-1\right)^2+r^2$$
A: 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
