**Factorization** is the decomposition of an expression into a product of its factors.

The following are common factorizations.

For any positive integer \(n\), \[a^n-b^n = (a-b)(a^{n-1} + a^{n-2} b + \ldots + ab^{n-2} + b^{n-1} ).\] In particular, for \( n=2\), we have \( a^2-b^2=(a-b)(a+b)\).

For \( n\) an odd positive integer, \[ a^n+b^n = (a+b)(a^{n-1} - a^{n-2} b + \ldots - ab^{n-2} + b^{n-1} ).\]

\( a^2 \pm 2ab + b^2 = (a\pm b)^2\)

\( x^3 + y^3 + z^3 - 3 xyz = (x+y+z) (x^2+y^2+z^2-xy-yz-zx)\)

\( (ax+by)^2 + (ay-bx)^2 = (a^2+b^2)(x^2+y^2)\). \( (ax-by)^2 - (ay-bx)^2 = (a^2-b^2)(x^2-y^2)\).

\( x^2 y + y^2 z + z^2 x + x^2 z + y^2 x + z^2 y +2xyz= (x+y)(y+z)(z+x)\).

Factorization often transforms an expression into a form that is more easily manipulated algebraically, that has easily recognizable solutions, and that gives rise to clearly defined relationships.

## 1. Find all ordered pairs of integer solutions \( (x,y)\) such that \(2^x+ 1 = y^2\).

Solution: We have \(2^x = y^2-1 = (y-1)(y+1)\). Since the factors \((y-1)\) and \((y+1)\) on the right hand side are integers whose product is a power of 2, both \((y-1)\) and \((y+1)\) must be powers of 2. Furthermore, their difference is

\[ (y+1)-(y-1)=2,\]

implying the factors must be \(y+1 = 4\) and \(y-1 = 2\). This gives \( y=3\), and thus \(x=3\). Therefore, \((3, 3)\) is the only solution.

## 2. Factorize the polynomial

\[f(a, b, c) = ab(a^2-b^2) + bc(b^2-c^2) + ca(c^2-a^2).\]

Solution: Observe that if \( a=b\), then \(f(a, a, c) =0\); if \(b=c\), then \(f(a, b, b)=0\); and if \( c=a\), then \( f(c,b,c)=0\). By the Remainder-Factor Theorem, \( (a-b), (b-c),\) and \( (c-a)\) are factors of \( f(a,b,c)\). This allows us to factorize

\[f(a,b,c) = -(a-b)(b-c)(c-a)(a+b+c).\]

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