1. Is there such number N that 7 divided N^2=3?

Isnt it just root of 7/3?


2. x^2 + y^2 = z^2. Prove xyz is a multiple of 60



Not sure what to do here, and where to get the xyz term from

1)

I assume you mean 7 divided by n^2 = 3

If you're dealing with divisibility in number theory, I think the question may be asking if there is an integer which divides 7 giving the quotient 3. This would mean that 7 would have 3 as a factor. 7, being prime, has no factors.

However the root of 7/3 does does yield a quotient of 3. The number does exist (root of 7/3), but it is not an integer.

2.

The solution set is . . Thus xyz is a multiple of 60.

so what about the set (5,12,13)?

You've got to try the following:
Prove that one of the numbers are divisble by 5,
prove that one of them is divisible by 3
and prove that one of them i divisible by 4.

These statements are true and should not be to hard to prove.

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(a²+b²)(c²+d²)=(ac+bd)²+(ad-bc)²