DISCRETE MATH - describe the range that belongs to n^2 + m^2 +1, where n and m are integers (no question mark)

Question

This is very difficult. At the moment I am writing down all the different numbers in the range from 0 up, which is showing no real pattern. This question is time-consuming and worth well over 10 points, but I would really appreciate any help.

Answer 1

Yes, this is a difficult question, so let's tackle it slowly.
Let m^2 + n^2 = t-1.
Then t-1 is representable as the sum of 2 squares
if and only if t-1 has no prime factor of the form 4k + 3
which occurs an odd number of times. This is a famous theorem
of Fermat. So if t is one more than such a number
it cannot be represented as m^2 + n^2 + 1.
The first few numbers not representable are t=
4, 7, 13, 15, 16, etc.
There's some information on the following website:
http://www.math.hmc.edu/funfacts/ffiles/20008.5.shtml

Answer 2

n = minus infinity to plus infinity
m= minus infinity to plus infinity
minus infinity to plus infinity is all set of real numbers including integers and decimals that can possibly exist. I think they're looking for an answer like "R".

If they didn't set that equation equal or less or greater than another expression or number, then any value can go within n and m without making the expression impossible. I don't see what other answer you can possibly give.


What level of math is that? I know Fermat's Theorem and I don't think the question is that hard. That equation is not set equal to anything so it's not about solving it (unless you left that part out). All they're asking is what are all the possible numbers that can go in that equation and if you look at it you can see that no matter what you put in for n and m you can find a solution to the problem. Something with 1/n^2 for example would be something different because 1 is not dividable by zero so it could not be "R".

You'd have to use Fermat's theorem if they gave you something like n^2 + m^2 =1 or equal to something. But that's not the case.