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any given number having any given power can be expressed as difference of infinite sets of two perfect squares like
5 ^ (3) = 63 ^ (2) - 62 ^ (2), 3 ^ (5) = 122 ^ (2) - 121 ^ (2) etc.

2006-06-12 03:05:30 · 3 answers · asked by rajesh bhowmick 2 in Science & Mathematics Mathematics

2 ^ (3) = (4.5) ^ (2) -(3.5) ^ (2) = 8

2006-06-12 04:14:17 · update #1

3 answers

See my answer in your next question, Rajesh.
Your answer only works for powers of 2. (2ⁿ)

And I agree with the other guy. (9/2)² does not yield a perfect square. What you're doing is taking 2^5 = 9^2 - 7^2 and dividing everything by 2^2 and saying you have a "new solution" to 2^3 = c²-b².

2006-06-12 07:42:40 · answer #1 · answered by bequalming 5 · 0 1

Clarify please.

8 = 2^3 = 9^2 - 1^2.
Would you show one of the other infinite pairs (m, n) such that 8 = m^2 - n^2?

Clarification 2:
Based on your answer, what is your definition of a perfect square? All the definitions that I have found require a perfect square to be the square of an integer.

2006-06-12 11:08:29 · answer #2 · answered by Anonymous · 0 0

any number N can be expressed as a^2-b^2

N=a^2-b^2 = (a+b)(a-b)=c.d

u can expressed any number as product of 2 other numbers c and d... this is obvious... i hope u dont need me to prove this.


to get a and b,

u have 2 linearly independent equations with 2 unknown:
a+b=c
a-b=d

solve equation to get a and b, then, voila, u have a and b.


p/s:
sorry. I didn't see "perfect". perfect square. no. not all numbers can be expressed as perfect squares. only those numbers which are sum of a few consecutive odd numbers can be expressed in this form. Or numbers which can be factored as c.d were c and d has to be both odd or both even.

2006-06-14 13:40:17 · answer #3 · answered by Anonymous · 0 0

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