#mathematics #polynomials #precalculus

Show that any odd number can be written as the difference of two squares.

Let the odd number be $$n = 2b + 1$$, where $$b$$ is a non-negative integer. Then we have

$n = 2b+1 = [ (b+1) + b ] [ (b+1) - b ] = (b+1)^2 - b^2. \ _\square$

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Difference Of Squares | Brilliant Math &amp; Science Wiki
wastes time and is, of course, boring. Notice that $$299=300-1$$ and $$301=300+1$$, so \begin{align*}299\times 301&=(300-1)(300+1)\\&=300^2-1^2\\&=89999. \ _\square \end{align*} <span>Show that any odd number can be written as the difference of two squares. Let the odd number be $$n = 2b + 1$$, where $$b$$ is a non-negative integer. Then we have $n = 2b+1 = [ (b+1) + b ] [ (b+1) - b ] = (b+1)^2 - b^2. \ _\square$ What is $234567^2-234557\times 234577\ ?$ Using the same method as the example above, \[\begin{align*} 234567^2-234557\times 234577&=234567^2-\big(234567^2-10^2\big)\\ &=23456