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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I also immediately found a solution: the two numbers can be 4 and 9.

It is the only solution.

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.

So the even is the square of a number ending in 2 or 8
and the odd is a square of a number ending in 3 or 7.

Now to check when the difference can be prime.  There are 4 combinations:
3²-2²=5
8²-3²=55
7²-2²=45
8²-7²=15
The difference of these squares 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.

Posted by Jer on 2014-11-14 09:02:55