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 re: All of the solutions
by Steve Herman)
for a^2-b^2 to be prime, a+b must be prime, and a-b=1, so we are looking for pairs of successive numbers a>b.
1st trial (2,1)==> check if 4+1 ends by 3 ; no good
2nd trial (3,2)==> check if 9+4 ends by 3 ; bingo...
easy to show lack of other solutions