answer this and become a genious in innovative mathematics.?

Question

take the equation
a ^ (n) - b ^ (n) = d ^ (2) - c ^ (2)
here a,b and n is given which are natural numbers.
find the mathematical relationship between a,b & d,c.
d and c can be rational numbers also.
hint :- its very simple.

7 ^ (3) - 4 ^ (3) = 48 ^ (2) - 45 ^ (2)

Answer 1

rajesh i donno what u mean but u have already provided the relation ship

Answer 2

If a + b is odd, there are two possible solutions for d and c.

One that satisfies this equation.

a^n - b^n = d^2 - c^2 = d + c

Where c is equal to the floor of (a^n - b^n)/2 and d is equal to the ceiling of (a^n - b^n)/2.

The other satisfies this equation.

a^n - b^n = d^2 - c^2 = 3 * (d + c)

Where c is equal to ((a^n - b^n)-3)/3 and d is equal to c+3.

If a + b is even, then there is only one solution for c and d.

d is then equal to [(a^n - b^n)/2 + 1/2] and c is equal to [(a^n - b^n)/2] - 1/2, or d - 1.

Answer 3

Ummmm, well there are many many solutions to this besides the one you gave. For instance, pick some integers a and b, let n be an even number, say n=2m, and then let d=a^m, c=b^m. Then
a^(n)-b^(n)=(a^m)^2-(b^m)^2=d^2-c^2

Answer 4

One easy answer:

Factorize,
a^n-b^n = (a-b)*(a^(n-1)+a^(n-2)*b+a^(n-3)*b^2+...+b^n).

Therefore (since c and d are allowed to be rational),

d-c = a-b
d+c = a^(n-1)+a^(n-2)*b+a^(n-3)*b^2+...+b^n

Or you can have
d+c = a-b
d-c = a^(n-1)+a^(n-2)*b+a^(n-3)*b^2+...+b^n.

So, d = sum/2; c = difference/2.

Answer 5

the above equation is correct when

a^n = b^n = d^2 = c^2

or

a^n = b^n and d^2 = c^2

0 is the ans in both sides of equation in this case

Answer 6

let a=19

b=10 and n=1

Then 19-10=(5)^2-(4)2

Answer 7

how can u deal with such bafflings??

Answer 8

IT IS EASY , BUT SORRY , I'M NOT GOING TO DO YOUR HOMEWORK FOR YOU .