The latex bragger is back!


From now on I am going to post the problem, solution with rendered latex, and bare solution.


We begin by seeing that both $15^6$ and $7^6$ have even powers, meaning that difference of squares is applicable.

From $a^2-b^2=(a+b)(a-b)$, we get $15^{3^{2}}-7^{3^{2}}=(15^3+7^3)(15^3-7^3)$

Another property in $a^3-b^3=(a-b)(a^2+ab+b^2)$ comes in handy to help further factorize the $(15^3-7^3)$ into $(15-7)(15^2+15\cdot 7+7^2)$
A similar property $a^3+b^3=(a + b)(a^2 - ab + b^2)$ can break apart $(15^3+7^3)$ into $(15+7)(15^2-15\cdot 7+7^2)$


Combining the two parts we end up with:
$$(15-7)(15^2+15\cdot 7+7^2)(15+7)(15^2-15\cdot 7+7^2)$$
$$(8)(225+105+49)(22)(225-105+49)$$
$$(8)(379)(22)(169)$$


Since $379$ is prime and also the largest number in the factorization, the largest prime factor of $15^6-7^6$ is $\boxed{\boxed{379}}$



Lyric(s) of the Post:I can't wait to leave this town
Cause lately I've been feeling down
The cold nights just don't feel the same

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