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.

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(In reply to

re: All of the solutions by Steve Herman)

even shorter:

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

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