1. ## Re: Four consecutive integers

$p=a^2-b^2=(a-b)(a+b)$

so $a-b$ is a divisor of $p$

Let $r=a-b$ and therefore $a+b=p/r$

solve for $a,b$

2. ## Re: Four consecutive integers

It seems you pick 4 natural numbers, it is quite 'likely' you will be able to express them as a difference of two squares:

How about 2x6x9x11 = 298^2-296^2 ?

Or

3x6x8x11 = 397^2-395^2 or 200^2-196^2 or 103^2-95^2

But i can't seem to find a way for this one...

3x6x9x11

What property will the product have than allows it to be expressed as a difference of two squares?

3. ## Re: Four consecutive integers

any odd integer can be written as the difference of two squares

$n=\left(\frac{n+1}{2}\right)^2-\left(\frac{n-1}{2}\right)^2$

$2^e=\left(2^{e-2}+1\right)^2-\left(2^{e-2}-1\right)^2$ if $e\geq 3$

The identity

$\left(a^2-b^2\right)\left(c^2-d^2\right)=(a c+ b d)^2-(a d+b c)^2$

says that the product of two integers, each a difference of two squares, is itself a difference of two squares

4. ## Re: Four consecutive integers

Wow, thank you.
I didn't know every odd integer can be expressed as difference of two squares. Of course, the product of four consecutive natural numbers will always be even.
So only some even numbers will not be expressible as a difference of two squares? How do we know 3x6x9x11 =1782 can't be expressed as a difference of two squares?

5. ## Re: Four consecutive integers

Originally Posted by rodders
How do we know 3x6x9x11 =1782 can't be expressed as a difference of two squares?
Suppose $a^2 - b^2 = 1782.$

Then a + b = larger factor of 1782
And a - b = smaller factor of 1782
2a = sum of the factors

But 1782 = 2*(odd number), so this is impossible.

For this to have a chance to work, the even number
must be able to be factored into two even factors.

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