Number Theory - Finding all possible triplets of two primes and one positive number Determine all positive integers a, b, c such that the numbers a² + 1 and b² + 1 are primes and the following equality 
(a² + 1)(b² + 1) = c² + 1
holds. 
My work: 
First experimentally found out one solution as a = 1, b = 2, c = 3. It satisfies the given equality with the associated constraints 
(1² + 1)(2² + 1) = 2*5 = 10 = 3² + 1. 
For problems involving the sum of two squares, I find that it is often useful to work in the ring of Gaussian integers Z[i]. In the present case we have the factorizations a² + 1² = (a + i)(a - i), b² + 1² = (b + i)(b - i), c² + 1² = (c + i)(c - i). The factors a + i, a - i, b + i, b - i are primes in the Gaussian integers, and each divides either c + i or c - i. Therefore, by using Gaussian Integers, Gaussian Primes, Existence of Prime Factorization, and Unique Factorization (by allowing unit multiples +/-1, +/-i), I could prove that is the one and only solution and so no other triplets exist which can satisfy the given requirements. 
But then the person who set the question liked my solution but remarked that "however there is a elementary solution." 
Could you please let me know about that if it can be solved using some other basic technique? 
High Regards,
Shamik Banerjee
 A: First observe that the LHS can be rewritten as $(ab+1)^2+(a-b)^2$ (and up to signs this is the only way it can be written as the sum of two squares of whole numbers.)
Now any prime larger than $3$ will require $a$ or $b$ to be even. So ...
A: Elementary solution:
We have:
$$a^2+b^2+a^2b^2=c^2$$
You ca use following identity which is derived from Pythagorean triples:
$$(2i+1)^2+[2i(i+1)]^2=[2i(i+1)+1]^2$$
Which gives:
$$(2i+1)^2+[2i(i+1)]^2+[(2i+1)^2.[2i(i+1)]^2=[2(i+1)(i+2)+1]^2$$
Which holds only when $i=1$.
But following identity holds for  $i=2$:
$$(2i+1)^2+[2i(i+1)]^2+[2(i+4)(i+5)]^2=[2(i+4)(i+5)+1]^2$$ 
$a=2i+1$ is odd, so $a^2+1$ can not be a prime, except when $i=0$ which gives $a^2+1=2$ .
$b=2i(i+1)$ is even, so $b^2+1$ can be a prime.
Example:$i=1$ results:
$$3^2+4^2+3^2.4^2=13^2$$
This is the only case where third term is of the form $a^2b^2$. as you can see :
$3^2+1=10$ is not prime, but $4^2+1=17$ is prime.
I could not find any more for $i ≥2$; for example $i=2$ by second identity gives:
$5^2+12^2+84^2=85^2$
Where $85≠5 \times 12$ 
Therefore what you found can be the only solution.May be better question is with this condition that $a^2+1$ or $b^2+1$ may be prime, then the number of solutions can be $2$ one is what you found second is what I showed. 
