Determine all triples $(x,y,p)$ of positive integers such that $p$ is prime, $p=x^2+1$ and $2p^2=y^2+1$.
1. We start with the given equations:
   \[
   p = x^2 + 1 \quad \text{and} \quad 2p^2 = y^2 + 1
   \]
2. Substitute \( p = x^2 + 1 \) into \( 2p^2 = y^2 + 1 \):
   \[
   2(x^2 + 1)^2 = y^2 + 1
   \]
3. Expand and simplify:
   \[
   2(x^2 + 1)^2 = 2x^4 + 4x^2 + 2 = y^2 + 1
   \]
   \[
   2x^4 + 4x^2 + 1 = y^2
   \]
4. Factorize the equation:
   \[
   (x^2 + 1)(2x^2 + 1) = y^2
   \]
5. Since \( p = x^2 + 1 \), we can rewrite the equation as:
   \[
   p(2x^2 + 1) = y^2
   \]
6. Consider the factorization \( (y - x)(y + x) = p(2x^2 + 1) \). We analyze two cases: \( p \mid (y - x) \) and \( p \mid (y + x) \).

### Case 1: \( p \mid (y - x) \)
7. Let \( y - x = pk \) for some positive integer \( k \):
   \[
   y = pk + x
   \]
8. Substitute \( y = pk + x \) into the factorization:
   \[
   p(2x^2 + 1) = (pk + x)(pk + x)
   \]
   \[
   p(2x^2 + 1) = pk^2 + 2xk + x^2
   \]
9. Simplify and solve for \( k \):
   \[
   2x^2 + 1 = k(pk + 2x)
   \]
   \[
   2x^2 + 1 = pk^2 + 2xk
   \]
   \[
   2x^2 + 1 = x^2k^2 + k^2 + 2xk
   \]
10. Since \( k \) must be an integer, we test \( k = 1 \):
    \[
    2x^2 + 1 = x^2 + 1 + 2x
    \]
    \[
    x^2 = 2x
    \]
    \[
    x(x - 2) = 0
    \]
    \[
    x = 2 \quad (\text{since } x \text{ is positive})
    \]
11. Substitute \( x = 2 \) back into \( p = x^2 + 1 \):
    \[
    p = 2^2 + 1 = 5
    \]
12. Substitute \( x = 2 \) and \( p = 5 \) into \( y = pk + x \):
    \[
    y = 5 \cdot 1 + 2 = 7
    \]
13. Thus, we have the solution \( (x, y, p) = (2, 7, 5) \).

### Case 2: \( p \mid (y + x) \)
14. Let \( y + x = pk \) for some positive integer \( k \):
    \[
    y = pk - x
    \]
15. Substitute \( y = pk - x \) into the factorization:
    \[
    p(2x^2 + 1) = (pk - x)(pk - x)
    \]
    \[
    p(2x^2 + 1) = pk^2 - 2xk + x^2
    \]
16. Simplify and solve for \( k \):
    \[
    2x^2 + 1 = k(pk - 2x)
    \]
    \[
    2x^2 + 1 = pk^2 - 2xk
    \]
    \[
    2x^2 + 1 = x^2k^2 + k^2 - 2xk
    \]
17. Since \( k \) must be an integer, we test \( k = 1 \):
    \[
    3x^2 + 1 = x^2k^2 + (k - x)^2
    \]
    \[
    1 = x^2(k^2 - 3) + (k - x)^2
    \]
18. If \( k \geq 2 \):
    \[
    1 \geq x^2 + (k - x)^2
    \]
    \[
    x = k = 1 \quad (\text{contradiction})
    \]
19. Thus, \( k = 1 \) does not yield a valid solution.

Conclusion:
\[
\boxed{(x, y, p) = (2, 7, 5) \text{ is the only solution}}
\]