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# What?

Find the largest prime factor of $$16^5 + 13^4 - 172^2$$. Any ideas?

Note by Sal Gard
3 months, 3 weeks ago

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I saw this same question on the AoPS forum (I believe it's from an ARML?), and realized no one has posted a solution, so if we're still interested,

$16^5 + 13^4 - 172^2 = 2^{20} + 169^2 - 172^2$

$S = 2^{20} + -3(341)$

$S = 2^{20} - 1023 = 2^{20} - 2^{10} + 1$

$S = \frac{2^{30} + 1}{2^{10} + 1}$

$S = \frac{1 + 4(2^7)^{4}}{1 + 2^{10}}$

We use Sophie-Germaine's identity to factorize $$a^4 + 4b^4 = (a^2 + 2b^2 + 2ab)(a^2 + 2b^2 - 2ab)$$ and thus,

$S = \frac{ (2^{15} - 2^8 + 1)(2^{15} + 2^8 + 1)}{2^{10} + 1}$

$S = 13 \cdot 61 \cdot 1321$ and $$1321$$ is prime. · 3 months, 2 weeks ago

Thanks for this great solution. Yes it was from the Arml in which I, for the first time, participated (I'm a sixth grader going into seventh.) I was trying to set up possible factors using Chinese remainder theorem systems. Are there any good approaches using this method? Either way, your solution is the most elegant. · 3 months, 1 week ago

At the end, how did you factorize S? I mean, for me it is not obvious that 1321 is a factor. · 3 months, 2 weeks ago

You can get that $$13$$ is a factor easily.
You can actually compute the terms in the numerator and note that $$2^{10} + 1 = 25 * 41$$ so you know what factors to look out for. · 3 months, 2 weeks ago

I point that out from the beginning. It is easier to compute the original expression... but that's not the case of the problem. · 3 months, 2 weeks ago

It may be easier to compute, but not easier to factorise. We already have split the numerator into two factors, and we also know $$25, 13$$ and $$41$$ are factors of their product - certainly that's progress?

Just for the sake of completeness, $\frac{2^{15} + 2^8 + 1}{25} = \frac{33025}{25} = 1321$ · 3 months, 2 weeks ago

Got it. Good solution! · 3 months, 2 weeks ago

What about the largest? I know there is a big one. · 3 months, 3 weeks ago