Let s = a 2-b 2 and t = c 2-d 2.

Then: st=(a+b)(a-b)(c+d)(c-d)

=(a+b)(c-d)(a-b)(c+d)

=[(AC-BD)+(BC-ad)][(AC-BD)-(BC-ad)]

=(ac-bd)^2-(bc-ad)^2

Because abcd is an integer, (ac-bd) and (bc-ad) must also be integers, obviously belonging to A.

Get a license.

The second question:

I tried to do it by mathematical induction, but I couldn't. What a pity, hehe

The third question:

Because they are all positive integers, there is a 1

Then you can only square the positive integer within 10.

It can be seen that A 1 = 1, A4 = 9, then a2 or a3 must be 3.

Because of their peace, 124

Then let's calculate; (Let's assume a2=3)

a 1+a4+a 1^2+a4^2+a2+a2^2= 104

Then the remaining two items are A3 and A3 2.

a3+a3^2=20

The solution is a3=4.

So a = {1, 3,4,9}, b = {1, 9, 16.438+0}