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I have to write $6^6 + 1$ as a product of prime numbers.

I don´t have any idea how to solve it based on arithmetical rules, so I computed the prime factors, which are 13, 37, and 97, algorithmically

I appreciate any hints

  • 10
    Hint: $(6^2)^3+1$ and polynomial $x^3+1$. – 2017-01-03
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    Hint $a^3+b^3=(a+b)(...)$ – 2017-01-03
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    Aside from the polynomial factoring, which works very well, you might note that if a prime $p$ divides your number then $6$ has order $12\pmod p$ so $12\,|\,p-1$. Narrows the search considerably. – 2017-01-03
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    @JeppeStigNielsen Now look, in a few minutes, I suspect 3 or 4 answers are going to show up, all basically the same due to your comment. :D – 2017-01-03
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    @lulu It's possible that $6^6\equiv -1\pmod {p}$ because $6^2\equiv -1\pmod p$. It's purely coincidental that $6^2+1\equiv 1\pmod{12}$ is a prime. But it is still a good technique for finding the primes other than the factors of $6^2+1$. – 2017-01-03
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    @ThomasAndrews Right you are. – 2017-01-03

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$6^6+1\\=(6^2)^3+1\\=36^3+1\\=(36+1)(36^2-36+1)\\=37 \times 1261$

$1261$ is divisible by $13$
$1261/13=97$

So, we get the prime factors
$37,13,97$

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    How do you know that 1261 is a multiple of 13 – 2017-01-03
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    using divisibility rule for $13$. see https://en.wikipedia.org/wiki/Divisibility_rule#Divisibility_by_13 and http://www.careerbless.com/calculators/divisibility/index.php – 2017-01-03
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    @AyoubFalah By comment by lulu (to the question, not to this answer), one would search $1\cdot 12+1=13$, (not $2\cdot 12+1=25$ which is composite), $3\cdot 12+1=37$, and so on. – 2017-01-03
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    @JeppeStigNielsen, won't applying divisibility rules step by step help for small numbers such as this? – 2017-01-03
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    @AyoubFalah For tests of divisibility for small primes, you can always add or subtract small multiples - so here, for example, notice that $1261+39=1300$ which makes the factor $13$ obvious. It is clear that $2,3,5,11$ are not factors. For $7$ you may know that $126$ is a multiple of $7$, so $1260$ is and $1261$ isn't. Here the trick is to find a known multiple of $7$ close to the target number. – 2017-01-03
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    Kiran, sure, but with lulu's observation we need only check [13, 37, 61, 73, 97, 109, 157, ...](http://oeis.org/A068228) which is nice. ___Edit:___ In fact, there must be a factor of $1261$ which does not exceed $\sqrt{1261}$, so it real has to be $13$. Cool. – 2017-01-03
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    ok, got the point – 2017-01-03
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$$(6^2)^3+1=(6^2+1)(6^4-6^2+1)\ldots$$

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I looked for possible factors $x$ such that $6^6\equiv -1 \bmod x$. Starting out it was immediately clear that some small numbers were excluded as factors; eg, $5$ ($6\equiv 1 \bmod 5$), $7$ ($6\equiv -1 \bmod 7$), $11$ ($\phi(11)=2\cdot5$)$

In particular I was looking for cases where $\phi(x)=12k$, and that pretty immediately gave me $13$ as a factor and $37$ as a factor. This meant the other factor was in the region of $100$, so a simple division test found $97$.

Bonus fact: since we have $13\mathord-1 \mid 6^6$, $37\mathord-1\mid 6^6$, and $97\mathord-1 \mid 6^6$, we also have $6^6+1$ is a Carmichael number.

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    The deduction that any prime factor $p$ must have $p-1$ divisible by $12$ is wrong in general. For example, that $12^6+1$ is divisible by $5$ and $29$. All the *other* prime factors are congruent to $1$ modulo $12$. Basically, the prime factors or $a^6+1$ are the prime factors of $a^2+1$ and the *rest* of the prime factors must have $p-1$ divisible by $12$. In this case, it is sort of an accident that $6^2+1$ is prime. – 2017-01-04
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    Yes, I realized that this is not a guaranteed property.However once I eliminated the possibility that $6^2+1$ has factors - which happened when $5$ was eliminated - I think I was safe. – 2017-01-04