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#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 & 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

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

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