Find my two numbers:
1. Both are positive integers..
2. Their difference is a prime.
3. Their product is a perfect square.
4. Their sum's last digit is 3.
Rem: It was solved by me immediately upon presentation by a friend.
I leave it to you to check whether there is more than one solution.
(In reply to All of the solutions
Nicely done, but your proof can be simplified further, as follows:
From 4. The numbers must be an even and an odd.
So to fit 3. The numbers themselves must be perfect squares.
All even perfect squares end in 0, 4, or 6.
All odd perfect squares end in 1, 5, or 9.
The only combination whose sum ends in 3 is 4+9.
Since the squares end in 4 and 9, their difference must end in 5.
The only prime that ends in 5 is 5 itself.
The only squares that are 5 apart are 4 and 9.
Hence, this is the only solution.