Decomposition to squares?

Question

Prove that
a^(n) - b^(n) can be expressed as difference of two squares in the form of (c+a-b)^(2) - c^(2), if a^(n) - b^(n) is not of the form of 4t + 2 where t is a rational number.Here "c" is natural number.

Answer 1

the basic idea is that :

an odd number can be easily expressed as the difference of two squares... ( m = ((m+1)/2)^2 - ((m-1)/2)^2 )

it is not so with even numbers... even numbers of the forms 4x can be expressed as the difference of two squares... (eg: 32 = 6^2 - 2^2) ; this can be easily proved... [ if (p^2-q^2) is even, then both (p+q) and (p-q) have to be even...hence it is a multiple of 4 ]

since 4t+2 is an even number that is not a multiple of 4, this cannot be expressed as the difference of two squares....for the same reason explained above (cannot be split into product of two even numbers)

there are better proofs.... but i think this hits the point more directly!

Answer 2

a^n-b^n is divisible by (a-b)

this is true because let f(a) = a^n-b^n
then f(b) = b^n-b^n = 0

so b is a zero of f(a).
so a-b is a factor

now (c+(a-b))^2 - c^2 = (c+a-b+c)(c+a-b-c)
= (2c+a-b))(a-b)
so (a-b) is a factor.

now we have to prove that other factor is more than this by even which will prove that c is natural number.

there are 2 cases

a and b both are odd
a^n is of the form 4n+1
b^n is of the form 4m+1
a^n-b^n dvisisible by 4
(a-b) is even if c is odd then given expression is odd so c is even

similarly we can prove when a is even b is odd
or both even
or a is odd b even

I think you mean t to be natural number

Answer 3

sorry I cant