Find all the pairs of positive numbers such that the last
digit of their sum is 3, their difference is a prime number and
their product is a perfect square.
(In reply to
solution by Larry)
Eliminate the last set, 'd', because whichever is the larger number, the difference will always be 5, and no primes end in 5.
Except the prime 5 itself can be said to end in 5. Then a solution (4,9) is valid 4+9=13 which ends in 3, 9-4=5 which is prime, and 4*9=36=6^2 which is a square.
re: solution
