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.

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.

@Mateo Matijasevick
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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,
\[ \frac{2^{15} + 2^8 + 1}{25} = \frac{33025}{25} = 1321 \]

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TopNewestI 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.

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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.

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At the end, how did you factorize S? I mean, for me it is not obvious that 1321 is a factor.

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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.

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Just for the sake of completeness, \[ \frac{2^{15} + 2^8 + 1}{25} = \frac{33025}{25} = 1321 \]

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What about the largest? I know there is a big one.

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Clearly 13 is a prime factor... does it help anything?

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