Express $4^8+6^8+9^8$ as a product of 2 numbers, each greater than 100.

Question

I attempted to factor it into $(4^4+3^4)^2$, but that gives a product of 2 times what is needed. Am I supposed to factor it, and how?

Answer 1

You are very close. You found (except for a typo) $$(4^4 + 9^4)^2 = 4^8 + 2 \cdot 36^4 + 9^8 = 4^8 + 2 \cdot 6^8 + 9^8.$$ All you need to do is subtract the extra power of $6$ to get $$ (4^4 + 9^4)^2 - 6^8 = 4^8 + 6^8 + 9^8.$$ Do you see how to factor it? Spoiler below is hidden.

The left-hand side can be factored as a different of squares.

Answer 2

Hint:

$$a^4+a^2b^2+b^4=(a^2+ab+b^2)(a^2-ab+b^2).$$

Answer 3

$$4^8+6^8+9^8=$$ $$(8113\times 5521) =$$ $$(133\times 336781)=$$

$$(427\times 104899) =$$ $$ (1159\times 38647)$$