Only solution to Diophantine equation with primes?

Question

Find all solutions to:

$$p_1 ^2 - p_2^2 = p_3^2 + 1$$

where $p_i$ is a prime number (positive of-course) The only solution that seems to exist is $(3,2,2)$ for $(p_1, p_2, p_3)$ accordingly. But I have no clue how to prove that no other solutions exist.. The only thing that jumps to my mind is assuming that another solution exists and somehow get to a contradiction...

I would highly appreciate your help, thank you!

Answer 1

Let $(p_1,p_2,p_3)\ne(3,2,2)$ If $p_1=2$ then we have $p_2^2+p_3^2=3$ which is impossible by the known theorem on the representation of a natural integer as a sum of two squares. Thus $p_1$ must be odd but in this case we have $p_2^2+p_3^2\equiv0\pmod4$ which is impossible since $p_2$ or $p_3$ must be distinct of $2$.

Answer 2

The square of a prime can only be $0$ or $1$ modulo $4$; the difference of two such squares cannot be $2$ modulo $4$. However, the RHS is $2$ modulo $4$ if $p_3$ is an odd prime. Thus $p_3=2$; this implies (again by the modular argument) that $p_2=2$, hence $p_1=3$ and the only solution is $3^2-2^2=2^2+1$.

Answer 3

Taken mod $8$, the squares of primes leave remainders $4$ (for $p=2$) and $1$ (for all odd primes). The only combination of those two remainders that satisfy the congruence $a-b\equiv c+1$ mod $8$ is $1-4\equiv4+1$ mod $8$, which means we need $p_2=p_3=2$, which in turns implies $p_1=3$.

Answer 4

Continuation of my comment and an alternative solution, from FLT $$3 \mid p^2 -1$$ for $p \ne 3$. Then check $$ p_1^2 - p_2^2=p_3^2+1 \tag{1}$$ $$\iff p_1^2 -1 - p_2^2+1=p_3^2-1+2$$ This will lead to $3 \mid 2$, if none of $p_1,p_2$ and $p_3$ is $3$. Thus, at least one of $p_1,p_2$ or $p_3$ must be $3$.

• if $p_1=3$ then $9-p_2^2=p_3^2+1$ yields $p_2=2$, since any other value will make $9-p_2^2\leq 0$, whether the RHS of $(1)$ is always positive.
• for this case, either $p_2=3$ or $p_3=3$ doesn't matter, we can swap them. So we have $$(p_1-p_2)(p_1+p_2)=p_1^2-p_2^2=10=2\cdot5$$ just 2 case to analyse $p_1-p_2=1,p_1+p_2=10$ and $p_1-p_2=2,p_1+p_2=5$.