aporter1 said:
so I'm thinking this is how i start,
r= (a2+b2)(c2+d2)...n=(ac-bd)2+(ad+bc)2
am i on the right track?
No. "n" in the way you used it *usually* refers to the n-th positive integer (i.e, the number "n" when you're counting 1,2,3,4,...n).
In a mathematical induction proof, the inductive step always involves this "IF it's (assumed) true for some n, THEN it can be proven true for (n+1)" step. You might be confused into thinking this means you're proving a property for some integer (n+1) - but this is not so. "n" is just an index of a general term a_n, and the exact meaning of the terms can vary depending on the problem. So if you're proving something about all positive integers, then a_n = n. If you're proving something about positive even integers, then a_n = 2n. And if you're proving something about prime numbers, then a_n is the nth prime number. And so forth.
In this problem, you're proving something about numbers that are sums of squares, so a_n simply refers to the n-th number on a list that's made up of numbers that are all sums of squares. The slightly confusing thing about this problem is that it may not be possible to order a_1, a_2, a_3,...,a_n consecutively, but it really doesn't matter. The subscripts are only used to distinguish the terms in the sequence.
[So here, n is being used in the context of a subscript (like a counting index) in the product of numbers, each of which is called "a" (just like SammyS used it). In my post, I used "k" as the subscript (and "n" to mean something else entirely), but here, I'll stick to SammyS's meaning of "n".]
What you want to write down is a product of certain numbers (called "a" with various subscripts) that are all individually sums of squares. Note that a_1, a_2, a_3,...,a_n are not necessarily consecutive integers, and they may not even be increasing in one direction. It doesn't matter. But the way you express that product (I'm calling the product big "A") is:
A_n = a_1a_2a_3...a_n.
In mathematical induction, you start with an assumption. Here the assumption is that A_n can be written as the sum of two squares. Let's call those squares p^2 and q^2, where p and q are integers. So you assume this to be true:
A_n = p^2 + q^2
What you now *need to prove* is that if you take any other integer (call it a_{n+1}) that's *also* the sum of two squares and multiply it by A_n, you can express that product as the sum of two squares also.
Because a_{n+1} can be written as s^2 + t^2, where s and t are integers, the product:
A_na_{n+1} = (p^2 + q^2)(s^2 + t^2)
Can you now use the identity given to express the RHS (right hand side) of that equation as the sum of two squares? Remember that "a" in the identity refers to something else - just another variable (it's unfortunate that they used the same symbol, but what can you do?).
Once you've done this, you've proved that assuming A_n can be written as the sum of two squares, the product A_{n+1} which is A_na_{n+1}, is also so expressible. That completes the inductive step.
The base step just involves a verification that the proposition (what you're trying to prove) holds for n = 1. And as I mentioned, this part is trivial, because the product here is just a_1, which is already defined to be the sum of two squares.
SammyS suggested a base step of n = 2, which involves a little algebra to prove that a_1a_2 is expressible as the sum of two squares. Frankly, I think this is unnecessary, but you can do it this way if you wish.
