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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 178

A number is called a perfect square if it is the square of an integer. How many pairs of perfect squares differ by 495? (Order does not matter. So, the pair "16 and 9" is the same as "9 and 16".)

I see you have graph paper.

You must be plotting something

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 93,650

Hi;

You did not say only positive integers so

are all solutions of x^2 - y^2 = 495

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 178

Exactly how many are there, I did 24, but it is wrong..

I see you have graph paper.

You must be plotting something

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 93,650

You did not say positive numbers or not. There are 24 positive and negative.

Only 6, just positive.

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

Considering just positive integers, there are 6 factorizations of 495 into two factors:

1*495, 3*165, 5*99, 9*55, 11*45 and 15*33. Each of these corresponds to one of

the six (x,y) pairs bobbym listed in post #4. For example: 1*495=248^2-247^2 and

24^2-9^2 = (24+9)(24-9) = 33*15.

In general for an odd composite number each of its unique factorizations (other than a perfect

square factorization) corresponds to a difference of squares. For 9 = 1*9 we get 5^2-4^2

= (5+4)(5-4) = 9*1 but 3*3 has no difference of squares representation unless we allow zero:

3^2-0^2 = (3+0)(3-0) = 3*3.

But we were talking about POSITIVE integers.

If M is an odd composite number and M=n*m where n and m are different, we get

as a difference of squares factorization.

If I recall correctly this was involved in one of Fermat's methods of factoring odd composites.

Have a grrreeeeaaaaaaat day!

Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).

LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

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