Perfect squares?

Question

Some natural numbers can be expressed as a difference of two squares, but others cannot. For example 12=4^2 - 2^2, but 10 cannot be written as a difference of two squares. I need a way to determine if a number can be determined by a difference of squares.

Answer 1

Any odd number can be expressed as the difference of two squares.

Any multiple of 4 can be expressed as the difference of two squares.

However, any even number that is NOT a multiple of 4 can not be expressed as the difference of two squares.


By the way, it's actually not too hard to prove this:

For any odd number: Write the number as 2k + 1. Then (k + 1)^2 - k^2 = 2k + 1. For example, for the number 7, k = 3. So (3 + 1)^2 - 3^2 = 7.

For a multiple of 4, write the number as 4k. Then (k + 1)^2 - (k - 1)^2 = 4k. For example, for the number 20, k = 5. So (5 + 1)^2 - (5 - 1)^2 = 20.

Finally, to prove the difference can't be an even number that is not a multiple of 4: Write a^2 - b^2 as (a + b)(a - b). Note that if a and b are both even or both odd, then (a + b) and (a - b) will both be even, therefore the product has to be a multiple of 4. However, if either a or b, but not both, are even, then (a + b) and (a - b) will both be odd, therefore the product has to be odd.

Answer 2

a^2-b^2=(a+b)(a-b) so for a number to be a difference of 2 squares, it must be factorable into that format.

12=(4-2)(4+2)

10*5*2. there is no way to express this as (a+b)(a-b) where a & b are integers. if you aren't limites to integers 10=3.5^2-1.5^2=12.25-2.25

if a number is prime, it can't be expressed as a^2-b^2.
if it is divisible by 2 but not 4, it can't be expressed as the difference of the squares of integers. Otherwise, I think they can.

Answer 3

The rule for this is actually
x2-y2=x+y ONLY when x and y are consecutive numbers.
for example 11squared-10squared = 12 = 10+11
So maybe if you want to determine if a number is the difference of two squares then try splitting it to see if is the sum of two consecutive numbers.

Answer 4

Let n be the number, a and b are two integers.
n = (a+b)(a-b)

Factor n into two integers l and m such that
a+b = l......(1)
a-b = m......(2)
will have integer solutution(s).

(1)+(2):
a = (l+m)/2
b = (l-m)/2
If n is an odd number, then l and m are both odd. Therefore, you can always find integers a and b such that n = a^2-b^2.

Answer 5

If it's odd, you write 2n+1=(n+1)^2 - n^2.
If it's even, it must be a mutiple of 4. So you write it as 4p and it is
(2p+1^2-(2p-1)^2; The only trouble is when it's of the form 4p+2, where you can't do it.