\[ S = 13 \cdot 61 \cdot 1321 \]
and \( 1321 \) is prime.
–
Ameya Daigavane
·
6 months ago

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@Ameya Daigavane
–
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.
–
Sal Gard
·
5 months, 3 weeks ago

@Mateo Matijasevick
–
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.
–
Ameya Daigavane
·
6 months ago

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@Ameya Daigavane
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I point that out from the beginning. It is easier to compute the original expression... but that's not the case of the problem.
–
Mateo Matijasevick
·
6 months ago

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@Mateo Matijasevick
–
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,
\[ \frac{2^{15} + 2^8 + 1}{25} = \frac{33025}{25} = 1321 \]
–
Ameya Daigavane
·
6 months ago

## Comments

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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. – Ameya Daigavane · 6 months ago

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– Sal Gard · 5 months, 3 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.Log in to reply

– Mateo Matijasevick · 6 months ago

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

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. – Ameya Daigavane · 6 months ago

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– Mateo Matijasevick · 6 months 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.Log in to reply

Just for the sake of completeness, \[ \frac{2^{15} + 2^8 + 1}{25} = \frac{33025}{25} = 1321 \] – Ameya Daigavane · 6 months ago

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– Mateo Matijasevick · 6 months ago

Got it. Good solution!Log in to reply

What about the largest? I know there is a big one. – Sal Gard · 6 months ago

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Clearly 13 is a prime factor... does it help anything? – Mateo Matijasevick · 6 months ago

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