Prove it with guts?

Question

In sir Fermats last theorem's equation
a^(n) + b^(n) = c^(n)
What will be "c" i.e. odd or even?I say it will be always odd .What do u say & why?

Answer 1

There is no solution for n > 2
for n =1 it is very trivial

so only one case remains when n =2

let us look for primitive roots

if c is even then

c^2 mod 4 = 0

but a^2 mod 4 = 0 or 1
b^2 mod 4 = 0 or 1

so a^2 mod 4 = b^2 mod 4 = 0

so b a and c have 4 as common so all are even so 2 can be factroed out

but if a is odd and b is even c cannot be even

if a is odd and b is odd then a^2+b^2 mod 4 = 2 so cannot be perfect square

so c shall always be odd(primitive root)



so

Answer 2

First of all it is not Sir Fernat (you meant Fermat) since he was never knighted by a king or queen of England. The proper salutation, for a French male would be Monsieur Fermat or Monsieur Pierre de Fermat. or simply Fermat.
Second here is the proof:
a) A square of an odd number is odd as a square of an even number is even.
b) A sum of an odd number and an even number is an odd number.
c) A sum of an even number and an even number is an even number and so is a sum of two odd numbers.

So c is odd iff a or b is odd. Iff they are both even or both odd then c is even.

Answer 3

example:

3^2+4^2=5^2 ,c is an odd no.

6^2+8^2=10^2 ,Cis not an odd no.

therefore,it depends on the a&b
if both are even then as in example it will be even no.

Answer 4

Even , if it is odd then we cam make it even
Suppose c is odd, the we can make a solution with c is even, by multiplying the a,b,c with a^n.

If c is even then either a and b are both even or both odd.

Answer 5

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