To find all triples \((p, q, r)\) of primes such that \(pq = r + 1\) and \(2(p^2 + q^2) = r^2 + 1\), we can follow these steps:
### Step 1: Analyze the equations
We have two equations:
1. \(pq = r + 1\)
2. \(2(p^2 + q^2) = r^2 + 1\)
### Step 2: Consider the parity of \(p\) and \(q\)
Since \(p\) and \(q\) are both primes, they can either be 2 (the only even prime) or odd primes.
- If both \(p\) and \(q\) are odd, then \(pq\) is odd, and thus \(r + 1\) is odd, which implies \(r\) is even. The only even prime is 2. Therefore, \(r = 2\). However, \(pq\) must be at least \(3 \times 3 = 9\), which gives \(r + 1 = 3\). This leads to a contradiction since \(r\) cannot be 2.
Thus, at least one of \(p\) or \(q\) must be 2.
### Step 3: Assume \(p = 2\)
Let’s assume \(p = 2\):
- Then from the first equation, we have:
\[
2q = r + 1 \implies r = 2q - 1
\]
- Substitute \(r\) into the second equation:
\[
2(2^2 + q^2) = (2q - 1)^2 + 1
\]
Simplifying this gives:
\[
2(4 + q^2) = (4q^2 - 4q + 1) + 1
\]
\[
8 + 2q^2 = 4q^2 - 4q + 2
\]
Rearranging the terms:
\[
0 = 2q^2 - 4q - 6
\]
Dividing the entire equation by 2:
\[
0 = q^2 - 2q - 3
\]
### Step 4: Solve the quadratic equation
Factoring the quadratic:
\[
(q - 3)(q + 1) = 0
\]
Thus, \(q = 3\) or \(q = -1\). Since \(q\) must be a prime, we have \(q = 3\).
### Step 5: Find \(r\)
Now substituting \(q = 3\) back into the equation for \(r\):
\[
r = 2(3) - 1 = 6 - 1 = 5
\]
### Step 6: Verify the solutions
We have found one solution:
\((p, q, r) = (2, 3, 5)\).
### Step 7: Assume \(q = 2\)
Now, let’s assume \(q = 2\):
- Then from the first equation:
\[
pq = r + 1 \implies p(2) = r + 1 \implies r = 2p - 1
\]
- Substitute \(r\) into the second equation:
\[
2(p^2 + 2^2) = (2p - 1)^2 + 1
\]
Simplifying gives:
\[
2(p^2 + 4) = (4p^2 - 4p + 1) + 1
\]
\[
2p^2 + 8 = 4p^2 - 4p + 2
\]
Rearranging:
\[
0 = 2p^2 - 4p - 6
\]
Dividing by 2:
\[
0 = p^2 - 2p - 3
\]
### Step 8: Solve the quadratic equation
Factoring gives:
\[
(p - 3)(p + 1) = 0
\]
Thus, \(p = 3\) or \(p = -1\). Since \(p\) must be prime, we have \(p = 3\).
### Step 9: Find \(r\)
Substituting \(p = 3\) back into the equation for \(r\):
\[
r = 2(3) - 1 = 6 - 1 = 5
\]
### Final Solutions
Thus, we have two valid triples:
1. \((p, q, r) = (2, 3, 5)\)
2. \((p, q, r) = (3, 2, 5)\)
