proof on difference of two squares and odd integers

Question

How would you prove that every odd integer is a difference of two squares?

I re phrased the problem to make it clearer to me: If $k$ is an odd integer then it ca be expressed in the form $a^2-b^2$ where a and b are integers.

So i start by supposing that k is an odd integer and so $k=2l+1$ for some l in the integers. It seems like you need cases for this but im not really sure how to do it. Could someone help me out please, thanks.

Answer 1

Hint: As you mentioned, you have $k = 2l+1$ for some $l$.

From here, note the difference of squares:

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

Let $a$ and $b$ be two consecutive integers and try to simplify.

Answer 2

Hint:

Compute $(n+1)^2-n^2$.

You should get $2n+1$

Answer 3

HINT

$$(k+1)^2-k^2=2k+1$$ $$k^2-(k+1)^2=-(2k+1)$$

Answer 4

Hint: What is the difference of squares of two consecutive numbers?

Answer 5

$$3=4-1$$

$$2n+1=(n+1)^2-n^2$$