Prove or disprove that $$7^{8}+8^{9}+9^{7}+1$$ is a prime number, without using a computer.
I tried to transform $n^{n+1}+(n+1)^{n+2}+(n+2)^{n}+1$, unsuccessfully, no useful conclusion.
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Bart Michels
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VividD
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6Checking mod 47 seems to work; there may be an easier way. – Adam Hughes Jul 16 '14 at 05:23
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3possible duplicate of Is $2013^{2014}+2014^{2015}+2015^{2013}+1$ a prime? (usage of a computer not allowed) – deostroll Jul 16 '14 at 05:25
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1Checking primes less than sqrt of this number though for divisibility doesn't ensure a quick solution. – Sidharth Ghoshal Jul 16 '14 at 05:28
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@deostroll, it is not a duplicate, different proof strategy should be employed... – VividD Jul 16 '14 at 05:28
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1whoever marked this question as a duplicate, spoiled the whole point of the question! connected two questions that are similar only in form... disaster... hope people think more before announcing "duplicates"! – VividD Jul 16 '14 at 05:34
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1Why do you expect there to be a reasonable way to do this by hand? – Théophile Jul 16 '14 at 05:38
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@Théophile Intuition. – VividD Jul 16 '14 at 05:40
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5@VividD: Consider that to be one of the drawbacks to writing overly sparse questions. How can you expect anyone else to treat the two questions as dissimilar if you yourself don't put any effort into making them dissimilar? – Jul 16 '14 at 05:55
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Have you attempted to factor $1 + 8^9$ or $1+ 9^7$ These both have clean factorizations in fact $x^n + y^n$ for all odd N can be factorized over $Z$ I'm hoping somewhere along the line the number 47 pops out frequently enough to warrant the check mod 47 – Sidharth Ghoshal Jul 16 '14 at 06:38
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Related. Perhaps there is an elegant solution, since this question has shown up before. – Bart Michels Jul 23 '14 at 07:58
2 Answers
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Trial division mod $47$, as per the hint.
Firstly, for positive integer $a$, observe that $$a \equiv \overbrace{a \mod 50}^{\text{reduced residue}}+3\lfloor a/50\rfloor \pmod {47}.$$ This makes taking mod $47$s much easier.
We compute $7^2=49 \equiv 2 \pmod {47}$. So $7^8 \equiv 2^4 = 16 \pmod {47}$.
We compute $8^2=64 \equiv 14+3 = 17 \pmod {47}$. So $8^4 \equiv 17^2 = 289 \equiv 39+3 \times 5=54 \equiv 4+3=7 \pmod {47}.$ So $8^8 \equiv 7^2 = 49 \equiv 2 \pmod 7 \pmod {47}.$ So $8^9 \equiv 2 \times 8 = 16 \pmod {47}$.
I happen to have memorized that $9^3=729$ (I used to set my alarm to 7:29am because it is equal to $3^6$). So $9^3 \equiv 729 = 29+3 \times 14=71 \equiv 21+3=24 \pmod {47}$. So $9^6 \equiv 24^2=576 \equiv 26+3 \times 11=59 \equiv 9+3=12 \pmod {47}$. So $9^7=12 \times 9=108=8+3 \times 2=14 \pmod {47}$.
Finally $16+16+14+1=47 \equiv 0 \pmod {47}$.
Rebecca J. Stones
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6Yes but but what would motivate reduction mod $47$ unless one had just set out to reduce mod everything up to $\sqrt{N}$? – 2'5 9'2 Jul 16 '14 at 07:45
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Indeed, it's a cop out that way. (I didn't feel like trial dividing $14$ more times.) – Rebecca J. Stones Jul 16 '14 at 07:50
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Still, there is some elegance and resourcefulness in this proof, even though 47 seems to be "known". – VividD Jul 16 '14 at 08:45
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So now we know how to use modulo in case we know the answer. But what if it was a prime? – Lord_Gestalter Jul 16 '14 at 11:38
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Patiently perform trial division using the $1440$ primes $\leq \sqrt{n}$? It might take three years of Sundays though. – Rebecca J. Stones Jul 16 '14 at 11:51
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@VividD The factor $47$ can be verified in a half-minute of mental arithmetic - see my answer. – Bill Dubuque Jul 22 '14 at 20:37
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@RebeccaJ.Stones: The last line of that Wikipedia article ("Cole died alone") seems like vandalism to me. – RghtHndSd Jul 22 '14 at 20:57
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That $\,47\,$ is a factor can be verified by very easy mental arithmetic:
$\begin{eqnarray} {\rm mod}\ 47\!:\quad && 1+ \color{#c00}{7^8} +\, 8^9 +\ \color{#0a0}{9^7}\\ \equiv && 1+ \color{#c00}{2^4} + 2^{27} + \color{#0a0}{7\cdot 7^6\cdot 8^7}\quad {\rm by}\ \ \color{#0a0}{9\equiv 7\cdot 8},\ \ \color{#c00}{7^2\equiv 2}\\ \equiv&& 1+ 2^4 + 2^{27} + 7\cdot \color{#c00}{2^3}\cdot 2^{21}\\ \equiv&& 1+ 2^4 + 2^{4} +\ 7\cdot 2\,\equiv\, 0\quad {\rm by}\ \ \color{#c00}2^{23}\equiv (\color{#c00}{7^2})^{23}\equiv 7^{46}\equiv 1,\ \ \rm by\ little \ Fermat\\ \end{eqnarray}$
Bill Dubuque
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