Editing Extended Euclidean algorithm (section)

====Example====

For example, if the polynomial used to define the finite field GF(2<sup>8</sup>) is {{math|1=''p'' = ''x''<sup>8</sup> + ''x''<sup>4</sup> + ''x''<sup>3</sup> + ''x'' + 1}}, and {{math|1=''a'' = ''x''<sup>6</sup> + ''x''<sup>4</sup> + ''x'' + 1}} is the element whose inverse is desired, then performing the algorithm results in the computation described in the following table. Let us recall that in fields of order 2<sup>''n''</sup>, one has &minus;''z'' = ''z'' and ''z'' + ''z'' = 0 for every element ''z'' in the field). Since 1 is the only nonzero element of GF(2), the adjustment in the last line of the pseudocode is not needed.
{| class="wikitable"
|-
! step
! quotient 
! r, newr
! s, news
! t, newt
|-
!
|  
| {{math|1=''p'' = ''x''<sup>8</sup> + ''x''<sup>4</sup> + ''x''<sup>3</sup> + ''x'' + 1 }}
|1
|  0
|-
!
|  
| {{math|1=''a'' = ''x''<sup>6</sup> + ''x''<sup>4</sup> + ''x'' + 1}}
|0
|  1
|-
! 1
| {{math|1=''x''<sup>2</sup> + 1}}
| {{math|1=''x''<sup>2</sup> = ''p'' &minus; ''a'' (''x''<sup>2</sup> + 1)}}
|1
| {{math|1=''x''<sup>2</sup> + 1 = 0 &minus; 1 &middot; (''x''<sup>2</sup> + 1)}}
|-
! 2
| {{math|1=''x''<sup>4</sup> + ''x''<sup>2</sup>}}
| {{math|1=''x'' + 1 = ''a'' &minus; ''x''<sup>2</sup> (''x''<sup>4</sup> + ''x''<sup>2</sup>)}}
| {{math|1=''x''<sup>4</sup>+''x''<sup>2</sup> = 0 &minus; 1(''x''<sup>4</sup>+''x''<sup>2</sup>)}}
| {{math|1=''x''<sup>6</sup>  + ''x''<sup>2</sup> + 1 = 1 &minus; (''x''<sup>4</sup> + ''x''<sup>2</sup>) (''x''<sup>2</sup> + 1)}}
|-
! 3
| {{math|1=''x'' + 1}}
| {{math|1= 1 = ''x''<sup>2</sup> &minus; (''x'' + 1) (''x'' + 1)}}
| {{math|1=''x''<sup>5</sup>+''x''<sup>4</sup>+''x''<sup>3</sup>+''x''<sup>2</sup>+1 = 1 &minus; (''x'' +1)(''x''<sup>4</sup> + ''x''<sup>2</sup>)}}
| {{math|1=''x''<sup>7</sup> + ''x''<sup>6</sup> + ''x''<sup>3</sup> +  ''x'' = (''x''<sup>2</sup> + 1) &minus; (''x'' + 1) (''x''<sup>6</sup>  + ''x''<sup>2</sup> + 1)}}
|-
! 4
| {{math|1=''x'' + 1}}
| {{math|1=0 = (''x'' + 1) &minus; 1 × (''x'' + 1)}}
| {{math|1=''x''<sup>6</sup> + ''x''<sup>4</sup> + ''x'' + 1 = (''x''<sup>4</sup>+''x''<sup>2</sup>) &minus; (''x''+1)(''x''<sup>5</sup>+''x''<sup>4</sup>+''x''<sup>3</sup>+''x''<sup>2</sup>+1)}}
|  
|}

Thus, the inverse is {{math|''x''<sup>7</sup> + ''x''<sup>6</sup> + ''x''<sup>3</sup> + ''x''}}, as can be confirmed by [[finite field arithmetic|multiplying the two elements together]], and taking the remainder by {{mvar|p}} of the result.