Determine all the pairs $(a,b)$ of positive integers, such that all the following three conditions are satisfied:
1- $b>a$ and $b-a$ is a prime number
2- The last digit of the number $a+b$ is $3$
3- The number $ab$ is a square of an integer.
To determine all pairs \((a, b)\) of positive integers that satisfy the given conditions, we will analyze each condition step-by-step.

1. **Condition 1: \(b > a\) and \(b - a\) is a prime number**

   Let \(b = a + p\), where \(p\) is a prime number.

2. **Condition 2: The last digit of \(a + b\) is 3**

   Since \(b = a + p\), we have:
   \[
   a + b = a + (a + p) = 2a + p
   \]
   The last digit of \(2a + p\) must be 3.

3. **Condition 3: The number \(ab\) is a square of an integer**

   Let \(ab = k^2\) for some integer \(k\).

Now, we will analyze the two cases mentioned in the solution:

### Case 1: \((a, p) = p\)

If \((a, p) = p\), then \(a = px^2\) and \(b = py^2\) for some integers \(x\) and \(y\). Substituting these into the equation \(b = a + p\), we get:
\[
py^2 = px^2 + p \implies p(y^2 - x^2) = p \implies y^2 - x^2 = 1
\]
This implies:
\[
(y - x)(y + x) = 1
\]
The only integer solutions to this equation are \(y - x = 1\) and \(y + x = 1\). However, this leads to \(y = 1\) and \(x = 0\), which implies \(a = 0\), which is impossible since \(a\) must be a positive integer.

### Case 2: \((a, p) = 1\)

If \((a, p) = 1\), then \(a = x^2\) and \(b = y^2\) for some integers \(x\) and \(y\). Substituting these into the equation \(b = a + p\), we get:
\[
y^2 = x^2 + p \implies y^2 - x^2 = p \implies (y - x)(y + x) = p
\]
Since \(p\) is a prime number, the factors of \(p\) are \(1\) and \(p\). Therefore, we have:
\[
y - x = 1 \quad \text{and} \quad y + x = p
\]
Solving these equations, we get:
\[
2y = p + 1 \quad \text{and} \quad 2x = p - 1 \implies y = \frac{p + 1}{2} \quad \text{and} \quad x = \frac{p - 1}{2}
\]
Thus:
\[
a = x^2 = \left(\frac{p - 1}{2}\right)^2 \quad \text{and} \quad b = y^2 = \left(\frac{p + 1}{2}\right)^2
\]

Now, we need to check the second condition: the last digit of \(2a + p\) must be 3. Substituting \(a\) and \(b\) into this condition, we get:
\[
2a + p = 2\left(\frac{p - 1}{2}\right)^2 + p
\]
Simplifying, we get:
\[
2a + p = \frac{(p - 1)^2}{2} + p = \frac{p^2 - 2p + 1 + 2p}{2} = \frac{p^2 + 1}{2}
\]
The last digit of \(\frac{p^2 + 1}{2}\) must be 3. Therefore:
\[
\frac{p^2 + 1}{2} \equiv 3 \pmod{10}
\]
Multiplying both sides by 2, we get:
\[
p^2 + 1 \equiv 6 \pmod{10} \implies p^2 \equiv 5 \pmod{10}
\]
The only prime number \(p\) that satisfies this condition is \(p = 5\).

Substituting \(p = 5\) into the expressions for \(a\) and \(b\), we get:
\[
a = \left(\frac{5 - 1}{2}\right)^2 = 4 \quad \text{and} \quad b = \left(\frac{5 + 1}{2}\right)^2 = 9
\]

Thus, the pair \((a, b) = (4, 9)\) satisfies all the given conditions.

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