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As the title says we seek to

Find a prime that divides $14^7+14^2+1$

There is a caveat though. This was part of a contest for high-school students so undergraduate Number Theory tools such as modular arithmetic should be avoided if we wish to remain true to the spirit of the competition.

That said, I did verify-employing exactly such tools-that if $p$ is the said prime then $p\neq2,3,5,7$.

Probably some form of simplifying the expression is required to solve it but it eludes me.

EDIT

The initial formulation of this question was "find the smallest such prime". This is highly unlikely to be achieved by pen and paper on the timeframe of a contest. For more details on why one can look at @lulu's answer below.

As noted on the comments by @JyrkiLahtonen, a similar question was posted before.

See here as well for answers.

Bill Dubuque
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MathematicianByMistake
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3 Answers3

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We can factor the polynomial $$x^7+x^2+1=(x^2+x+1)\times (x^5-x^4+x^2-x+1)$$

Letting $x=14$ shows that $211$ is a prime factor. You still have to prove it is the least prime factor, but at least you can work with a smaller number.

lulu
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  • ...how do you know one of the sides is the smallest prime for $x=14$? – Peter Woolfitt Oct 27 '17 at 21:36
  • Ok, this simplifies things. +1 – MathematicianByMistake Oct 27 '17 at 21:37
  • @MathematicianByMistake How long did competitors have to solve the problem? – Peter Woolfitt Oct 27 '17 at 21:38
  • @PeterWoolfitt Because it is easy to check that $211$ is prime. You still have to prove that the other factor, $499591$, is not divisible by any prime $<211$ but at least this is a simpler computation. – lulu Oct 27 '17 at 21:38
  • @lulu the computation still seems infeasible for a high school math contest – Peter Woolfitt Oct 27 '17 at 21:40
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    @PeterWoolfitt There were $4$ questions-if I am not mistaken this was the last one (most difficult?) and participants had $3$ hours to finish. – MathematicianByMistake Oct 27 '17 at 21:41
  • @PeterWoolfitt It's pencil and paper. I grant you it is a bit ugly. Personally, I declined to check the other factor (up to $211$) by hand. Wouldn't take more than $15$ minutes though. Anyway, this is the best idea for it I have! – lulu Oct 27 '17 at 21:42
  • @MathematicianByMistake Oh wow, in that case, you could indeed make this method work. – Peter Woolfitt Oct 27 '17 at 21:42
  • Ok fellows, I messed up-the exact wording of the test-I did not have a chance to see it myself yet-was "find a prime number" ,not the smallest. I will obviously accept this answer and Edit the question. Sorry – MathematicianByMistake Oct 27 '17 at 21:44
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    @MathematicianByMistake Oh, that's much easier. Thanks for clarifying! – lulu Oct 27 '17 at 21:45
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    Nice, though I wonder how does one arrive at this factorization with pencil and paper :) – Sil Oct 27 '17 at 21:46
  • @lulu I just found out myself-a friend's pupil took part on this so I did not have a first-hand account. Thanks again for your answer! – MathematicianByMistake Oct 27 '17 at 21:47
  • How did you factor that polynomial? – Ovi Oct 27 '17 at 23:27
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    @Ovi It's essentially cyclotomic. Obviously a primitive cube root of $1$ is a root so it is clear that $x^2+x+1$ is a factor. Actually, that remark is sufficient to solve the problem, but once you have one factor it is easy to find another. This sort of factoring comes up all the time in high school math challenge problems. – lulu Oct 27 '17 at 23:37
3

The reason we can get one factor immediately is that $7 \equiv 1 \pmod 3.$ As a result, both nontrivial roots of unity are roots of $x^7 + x^2 + 1,$ meaning that $(x - \omega)(x- \omega^2) = x^2 + x + 1$ must divide the polynomial $x^7 + x^2 + 1.$ Here $\omega^3 = 1$ but $\omega \neq 1$

If this seems uncomfortable, just consider that $x^2 + x + 1$ is the minimal polynomial for $\omega$ over $\mathbb Q,$ and must divide any polynomial for which $\omega$ is a root. Furthermore, the Gauss theorem on content tells us that the quotient polynomial has integer coefficients, not just rational.

Similar: the polynomial $x^{141} + x^{93} + x^{82} + x^{44} + 1$ is divisible by $x^4 + x^3 + x^2 + x + 1,$ consider a fifth root of unity.

Will Jagy
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$$x^7+x^2+1=x^7-x+x^2+x+1=x(x^3-1)(x^3+1)+x^2+x+1=$$ $$=(x^2-x)(x^2+x+1)(x^3+1)+x^2+x+1=$$ $$=(x^2+x+1)(x^5-x^4+x^2-x+1).$$

Michael Rozenberg
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