Any power & the squares?

Question

Prove that for the equation
x^(n)=a^(2) - b^(2)
where "x" is any natural number with any natural power can be expressed as difference of two perfect squares "a" & "b".

Answer 1

if x is odd then

x^n is odd say 2n+1

2n+1 = (n+1)^2 - n^2

the it can be done

if x is even then x^n is divisible by 2^n

if n =1 or 2 this is not always true for example

2 is not difference of any 2 squares nor is 4


but for other n it is true

2^n = (a+b)(a-b)
we can factor 2^n into 2 distinct even numbers a+b and a-b from which a and b can be found out.

(if both are not distinct then a-b = a+b or b = 0)

Answer 2

If you divide a number by 4 and get a remainder of 2, you cannot express the number as the difference of two squares. If the remainder is 0,1,or 3, then you can.

(All squares of odd numbers give a remainder of 1 when divided by 4. All squares of even numbers give a remainder of 0 when divided by 4. So the difference of two odd squares or the difference of two even squares gives a remainder of 0 when divided by 4. An odd square minus an even square always gives a remainder of 1 and an even square minus an odd square always gives a remainder of 3 when they are divided by 4.)

A number is the difference of two squares iff it can be expressed as the product of two numbers of the same parity, i.e. both even or both odd.

( Let a = b^2 - c^2 . Then a = (b+c)(b-c), and b+c and b-c are always either both odd or both even. Conversely, if a = rs and r and s are both odd or both even, then r+s and r-s are always both even, so if b = (r+s)/2 and c = (r-s)/2 then b and c are both integers.
b^2 = (r^2 + 2rs + s^2)/4. c^2 = (r^2 - 2rs + s^2)/4.
b^2 - c^2 = (r^2 + 2rs + s^2 - r^2 + 2rs - s^2)/4 = 4rs/4 = rs = a.)

Answer 3

Hmmm. with regard to the 1st venture the place -2^2 = 4. i think of that's misguided, it would've been -(2)^2 = -4. If we've -2^2, it is comparable to (-2)^2 it somewhat is comparable to 4. To the 2nd venture, with 2^2 -16x = 0, a danger solutions are 0 and sixteen for x. If she pronounced that the respond is x = 0 or x = -sixteen , the equation might've been -x^2 - sixteen = 0. If x = -sixteen, then -(-sixteen)^2 - sixteen(-sixteen) = 0 - 256 + 256 = 0 0 = 0 undergo in recommendations that putting parentheses properly is quite substantial. . . appropriate me if i'm incorrect.

Answer 4

clearly,
x^n=(a+b)(a-b)..........................since a^2-b^2=(a+b) (a-b)
now, (a+b)(a-b) can form any natural number(even or odd)
now a,b are perfect squares
=>a,b belong to whole numbers
=>x^n can be expressed in as a^2-b^2 if and only iff a is not equal to b.

Answer 5

5^2-3^2=4^2
5^2-4^2=3^2
10^2-8^2=6^2
10^2-6^2=8^2