Recent content by robert Ihnot

This is a famous problem and not an easy one. Euler first proved it. I quote from Wikipedia: "Fermat's theorem on sums of two squares asserts that an odd prime number p can be expressed as p = x2 + y2 with integer x and y if and only if p is congruent to 1 (mod 4). The statement was...

When I took the Calculus, it was the difinition of e^x, that is as x goes to infinity. Ah, but so how did you manage to write [SIZE="3"]e^x = ∑ x^n/n! It popped out from nothing without using the Taylor expansion?

The matter turns on inverses. For every a in the system there is an a^-1, such that a*a^-1 = 1. (Generally that is an axiom of the group property.)

I did not think a physic's professor or an engineer an expert on a mathematical statement. All I was taking objection to was the statement: Jarel: That sin(x) approaches x as x approaches 0 doesn't make any sense. When I was working those pendulum problems in Physics years ago, we did...

No, that's not true. Our college physics professor put that it that way, sin(x) approaches x, as x is very small. It is a way of quickly calculating small values of sin(x). I believe he said for values under 5 degrees, transulated, of course, into radients. All this happened before the...

You seem to assume that everyone knows what you are talking about--why? You certainly have not defined terms, but when I look into this, it is my impression that you have no understanding of modular arithmetic. Why not check that out?

If induction goes over the entire set of natural numbers then you can call this the infinite case. Which is the same as for every natural number, i.e., we have a sucessor--namely n passes to n+1. Thus we may conclude that all natural numbers have been included in the set and such a set is...

How about 3? phi(3)=2, phi(2)=1.

Check me out page 2.

Another way to approach this whole thing depends upon using the Taylor Theorem, but in a very truncated form. I will give an example: F(x) = 2^5 + x^5 == 0. Mod 5. In this case, x=3. It is very easy to move to the case Modulus=25, which has already been show, first by lpetnich, and then...

The point of the situation with regards to p^2, is that, for example p=3, we have terms like 9!/(3!*6!) in the binominal expansion of (a+b)^9. In the mention term = 84, we have only 3, not 9, as a divisor. Fermat's Little Theorm can be shown by induction: 1^p ==1, (x+1)^p ==x^p+1, but...

Here is a case for mod 9. Now since phi of 9 = 6, we need only raise these numbers to the cube, but if you like, use 9 instead of 3. 2^3+5^3 \equiv133\equiv7 Mod 9. But 7^3\equiv 1 Mod 9.

I would be helpful to give a reference or show a proof.

Right above!

p^2 divides p^(a - b), which is congruent to p^a - p^b. By Fermat's Litle Theorm, Modulus p. Fermat's Little Theorem says nothing about modulus P^2.