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If you can solve this problem, then you are a math expert?
Pick two natural numbers that differ by 30, and their product is a perfect square. One example is 2, 32. Their product is 64 which is a perfect square. There are three more such possible pairs of numbers. Find the three pairs of numbers.
I'll give 10 points if you can find the three pairs of numbers, but I'll consider you as a math expert if you can go one step further and prove that there can be only a maximum of four such pairs of numbers which differ by 30 and their product is a perfect square.
Let hte numbers be n and n+30.
n*(n+30) = k^2
Case 1: n and (n+30) have all common factors
Possible factors are factors of 30 = 2, 3, 5, 6, 10, 15
From this (2, 32) and (10, 40) work.
Case 2: n and (n+30) both have a set of different factors with even multiplicity.
From this 24 (=2^2 * 6) and 54 (=3^2 * 6) work
Also (98, 128)
So far that's four: (2, 32); (10, 40); (24, 54); (98, 128)
****
Regarding the final part of your question, Jon has it nailed although he doesn't advertise it very well. His line
d=-15+-sqrt(15^2+(k/2)^2) should really be
d=-15+ sqrt(15^2+k^2)
with the conclusion that
15^2 + k^2 = perfect square = s^2
or 15^2 = 225 = s^2 - k^2 = difference between two perfect squares.
We know that each pair of consecutive perfect squares has a difference that follows the odd number sequence.
1,4,9,16
difference = 3, 5, 7, ...
That means there will be an upper limit to the possible values of k and s.
113^2 - 112^2 = 225
Any pair of perfect squares exceeding 113 will result in a RHS greater than 225, thus will not work. This corresponds to the (98, 128) solution. Therefore this is the largest pair of natural numbers satisfying your problem. Since there are only three solutions less than this, these must be the only four solutions.
***
On an entirely different note, why would someone with 9 points wish to keep their Q&A private?