Maths Problems Thread

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#9[difficult]
No three positive integers a, b, and c can satisfy the equation a^n + b^n = c^ n for any integer value of n greater than two. Give a proof to this theorem
[troll]

I doubt I can do Fermat's Thm but I can get a little way. If n=uv then (a^u)^v + (b^u)^v = (c^u)^v so there would be a solution to the case with n=v. Therefore we only need to prove for n an odd prime or 4. (This is a well-known idea and not originally mine.)

If gcd(a,b,c)>1 then a solution a' = a/g, b' = b/g, c' = c/g exists, hence we only have to prove a,b and c with gcd(a,b,c)=1. (Probably also not original.)

We need only prove for pairwise-coprime (a,b,c) since... if g := gcd(x,y) > 1 with x and y two different elements of (a,b,c) and z = the third element, then a^n + b^n = c^n (mod g) => z^n = 0 (mod g) => z=0 (z can't be any other zero divisor of g since then gcd(x,y,z)>1 which we proved above was a case we didn't need to consider). (Probably also not original.)

This means we only need to prove it for n=3, 4, 5, 7, 11, .. and pairwise coprime a, b and c.

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#1 solution:

Spoiler

Without loss of generality let's take a ≤ b < c.

If a=2 then b^2 = c^2-4 = (c-2)(c+2). We can assume c>3 since the case c=3 leads to b^2 = 5. So (c-2)(c+2) contains at least two different prime factors p and q, so p|b^2 => p|b and q|b^2 => q|b, hence b is not prime.

If a>2 then a,b and c are all odd so a^2+b^2=0≠1=c^2 (mod 2).

Spoiler

To see why (c-2)(c+2) has two different prime factors we note c must be odd since c has to be a prime greater than 3. Then gcd(c-2, c+2) = gcd(4, c-2) = 1 since c-2 is odd.

#2 solution:

Spoiler

There exists an integer b with b > 1/(x'-x) => 1/b < x'-x (α).

Consider the interval (bx, bx'), with width b(x'-x) > 1 (using α). So there must be an integer a in (bx, bx'). So we see bx < a < bx' => x < a/b < x' as required.