# if  a^2=2a+1 prove  a^5=29a+12

You need to use the following exponential identity such that:

`a^b = a^(b_1+b_2+...+b_n) = a^(b_1)*a^(b_2)*...*a^(b_n)`

Reasoning by analogy yields:

`a^5 = a^(2+2+1) => a^5 =a^2*a^2*a`

Notice that the problem provides that `a^2 = 2a + 1 => a^2 - 2a = 1`

Completing the square yields:

`a^2 - 2a + 1 = 1 + 1 => (a - 1)^2 = 2 => a - 1 = +-sqrt2`

`a = 1 +- sqrt2`

Substituting `1 +- sqrt2`  for a yields:

`a^5 = (1+sqrt2)^2*(1 + sqrt2)^2*(1 + sqrt2)`

`a^5 = (1 + 2sqrt2 + 2)^2*(1 + sqrt2)`

`a^5 = (3 + 2sqrt2)^2*(1 + sqrt2)`

`a^5 = (9 + 12sqrt2 + 8)(1 + sqrt2)`

`a^5 = (17 + 12sqrt2)(1 + sqrt2) => a^5 = 17 + 17sqrt2 + 12sqrt2 + 24`

`a^5 = 41 + 29sqrt2`

You need to write `41 = 29 + 12`  such that:

`a^5 = 12 + 29 + 29sqrt2`

Factoring out 29 yields:

`a^5 =12 + 29(1 + sqrt2)`

Notice that you may substitute back `a`  for `1 + sqrt2`  such that:

`a^5 = 12 + 29a`

Hence, evaluating `a^5` , using the given relation `a^2 = 2a + 1`  `yields a^5 = 12 + 29a` .

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