$p^2+1=q^2+r^2$. Strange phenomenon of primes

Question

Problem: Find prime solutions to the equation $p^2+1=q^2+r^2$

I welcome you to post your own solutions as well

I have found a strange solution which I can't understand why it works(or what's the math behind it.) Here it is through examples

Put $r=17$(prime) Now $17^2-1=16\times 18=288=2 \times 144$ (a particular factorization)

$\frac {2+144}{2}=73$

$\frac {144-2}{2}=71$

Solution pair $(p,q,r)=(73,71,17)$

Put $r=23$, $23^2-1=22\times 24=8\times 66$

$\frac {8+66}{2}=37$

$\frac {66-8}{2}=29$

Solution pair $(37,29,23)$

It works for each prime except for $2,3,5$ Which generate $(2,2,1),(3,3,1),5,5,1)$

Please explain me how it's working

Answer 1

This seems to hold much more broadly than just for primes.

Let$$r^2-1=ab$$where $b>a$, and let $p=\frac{b+a}{2}$ and $q=\frac{b-a}{2}$.

Then the equation $p^2+1=q^2+r^2$ becomes$$\left(\frac{b+a}{2}\right)^2+1=\left(\frac{b-a}{2}\right)^2+ab+1$$and we have$$\frac{b^2+2ab+a^2}{4}+1=\frac{b^2-2ab+a^2+4ab}{4}+1=\frac{b^2+2ab+a^2}{4}+1$$

Addendum: If $r$ is an odd prime, then $p$ and $q$, that is $\frac{b+a}{2}$ and $\frac{b-a}{2}$, can be prime only if both are odd, i.e. only if $b\equiv 2\pmod 4$ and $a\equiv 0\pmod 4$, or vice-versa.

This restricts the number of divisor pairs $a, b$ that need to be considered in determining whether $r^2-1=ab$ for some odd primes $p$ and $q$.

To illustrate, let $r=47$. Then$$r^2-1=ab=2208=2^5\cdot3\cdot23$$Thus the possible $ab$ are $2\cdot 1104$, $6\cdot 368$, and $2^4\cdot 138$.

For $b=1104$, $a=2$, $p=\frac{b+a}{2}=553=7\cdot79$, not prime.

For $b=368$, $a=6$, $p=\frac{b+a}{2}=187=11\cdot17$, not prime.

For $b=138$, $a=2^4$, $p=\frac{b+a}{2}=77=7\cdot11$, not prime.

It seems $47$ is the least prime value of $r>3$ for which$$p^2+1=q^2+r^2$$is true for no primes $p$, $q$.

As @Gerry Myerson shows, $r=193$ is another instance. I've yet to find other primes between these two, but have not looked much beyond $r=97$.

Answer 2

Let $r=193$, a prime. Then $$r^2-1=9313^2-9311^2=3107^2-3101^2=323^2-259^2$$ are the only expressions of $r^2-1$ as a difference of two squares, but $9313=67\times139$, and $3101$ and $259$ are both multiples of $7$. Thus, $p^2+1=q^2+r^2$ is impossible in primes $p,q$ for this prime value of $r$.

Answer 3

The below identity can be utilized to get prime solutions.

$a^2+1=b^2+c^2$

$(mp+nq)^2+(mn-pq)^2=(mn+pq)^2+(mp-nq)^2$ ---(1)

Since one of the four elements in equation (1) needs to be equal to one we take:

$(mn-pq)=+1 or -1$

We choose:

$(m,n,p,q)=(3,4,11,1)$

$mn-pq=12-11=1$

And we get: $(a,b,c)=(37,29,23)$

Also for, $(m,n,p,q)=(5,7,9,4)$

We get: $(mn-pq)=(35-36)=(-1)$

And, $(a,b,c)= (73,17,71)$

$(m,n,p,q)=(7,6,41,1)$

$(mn-pq)=(42-41)=1$

$(a,b,c)=(293,281,83)$