$36^8+1$

$$ Rewrite $$ $\begin{aligned} (36+1)(36^2+1)(36^4+1)(36^8+1)&=(6^2 + 1)(6^4 + 1)(6^8 + 1)(6^{16} + 1)\\ &=\frac{(6^4 - 1)}{(6^2 - 1)}\cdot\frac{(6^8 - 1)}{(6^4 - 1)}\cdot\frac{(6^{16} - 1)}{(6^8 - 1)}\cdot\frac{6^{32} - 1}{6^{16} - 1}\cdot\\ &=\frac{6^{32} - 1}{(6^2 - 1)}\\ &=\frac{(6^{32} - 1)}{35}. \end{aligned}$ $$ Thus, $a=32$ , $b=35$ , and $a+b=\boxed{67}$ . $$ $\text{\# }\mathbb{Q.E.D.}\text{ \#}$

We can use the equation $(a-b)(a+b) = a^{2} - b^{2}$ .First,we multiply both sides by $(36-1)=35$ to get

$(36-1)(36+1)(36^{2}+1)(36^{4}+1)(36^{8}+1) = \frac{6^{a}-1}{b} \times 35$

By multiplying the terms on the left hand side,we get

$(36^{2}-1)(36^{2}+1)(36^{4}+1)(36^{8}+1) = \frac{6^{a}-1}{b} \times 35$

$= (36^{4}-1)(36^{4}+1)(36^{8}+1) = \frac{6^{a}-1}{b} \times 35$

$= (36^{8}-1)(36^{8}+1) = \frac{6^{a}-1}{b} \times 35$

$= 36^{16}-1 = \frac{6^{a}-1}{b} \times 35$

Because $36 = 6^{2}$ , $36^{16} = (6^{2})^{16}$ which is equal to $6^{32}$ since $(6^{2})^{16}$ is equal to $6^{2} \times 6^{2} \times 6^{2} .........$ 16 times.So we have

$6^{32}-1 = \frac{6^{a}-1}{b} \times 35$

The factors of 35 are 1,5,7 and 35.Because $6^{32}-1$ is a positive integer , $b$ must be 1,5,7 or 35 for $\frac{6^{a}-1}{b} \times 35$ to be an integer.It is easy to observe that $\frac{6^{a}-1}{b} \times 35$ is an integer of the form $6^{a}-1$ only if $b = 35$ ,so we have $6^{32}-1 = \frac{6^{a}-1}{35} \times 35$ .After cancelling out the 35, we can find that $a = 32$ , so $a + b = 32 + 35 = 67$ .

This can be written as (6^2 + 1) * (6^4 + 1) * (6^8 + 1) * (6^16 + 1) =

[(6^4 - 1)/(6^2 - 1)] * [(6^8 - 1)/(6^4 - 1)] * [(6^16 - 1)/(6^8 - 1)] * [(6^32 - 1)/(6^16 - 1)] =

(6^32 - 1) / (6^2 - 1) = (6^32 - 1) / 35. Thus a = 32, b = 35 and so a + b = 67.

(36+1)(36^2+1)(36^4+1)(36^8+1)=(36^16-1)/(36-1)=(6^32-1)/35 hence a=32, b=35 & a+b=67