Find all the pairs of positive numbers such that the last
digit of their sum is 3, their difference is a primer number and
their product is a perfect square.
1. Let the numbers be \(a\) and \(b\).
2. Suppose \(\gcd(a, b) = d > 1\). Write \(a = da_1\) and \(b = db_1\), where \(\gcd(a_1, b_1) = 1\).
3. Given that their difference is a prime number, we have \(d(a_1 - b_1) = p\), where \(p\) is a prime number.
4. Given that the last digit of their sum is 3, we have \(d(a_1 + b_1) \equiv 3 \pmod{10}\).
5. Since \(d > 1\), we get that \(a_1 - b_1 = 1\). Thus, \(a_1 = b_1 + 1\).
6. Given that their product is a perfect square, we have \(a \cdot b = d^2 a_1 b_1 = x^2\) for some integer \(x\).
7. Substituting \(a_1 = b_1 + 1\) into the product condition, we get \(d^2 b_1 (b_1 + 1) = x^2\).
8. Since \(b_1\) and \(b_1 + 1\) are coprime, we have that \(b_1 = y^2\) and \(b_1 + 1 = z^2\), where \(y\) and \(z\) are positive integers.
9. This implies \((z - y)(z + y) = 1\). Since \(y, z \geq 1\), we have \(z + y \geq 2\) and \(z - y \in \mathbb{Z}\), which is a contradiction.
10. Thus, \(\gcd(a, b) = 1\). Therefore, we have that \(a = m^2\) and \(b = n^2\), where \(m\) and \(n\) are positive integers.
11. Let \(a > b\), meaning \(m > n\).
12. Therefore, \(m^2 - n^2 = p\), where \(p\) is a prime number.
13. Factorizing, we get \((m - n)(m + n) = p\). Thus, \(m + n = p\) and \(m - n = 1\).
14. Solving these equations, we get \(2m = p + 1\) and \(2n = p - 1\).
15. Given that their sum is \(\equiv 3 \pmod{10}\), we have \(\frac{p^2 + 1}{2} \equiv 3 \pmod{10}\).
16. By the Chinese Remainder Theorem (CRT), \(\frac{p^2 + 1}{2} \equiv 1 \pmod{2}\) and \(\frac{p^2 + 1}{2} \equiv 3 \pmod{5}\).
17. Hence, \(p^2 \equiv 0 \pmod{5}\), i.e., \(p = 5\).
18. We conclude that \(m = 3\) and \(n = 2\). Thus, \(a = 9\) and \(b = 4\).
19. The only pair is \((9, 4)\).

The final answer is \( \boxed{ (9, 4) } \).