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show that: there isn't exsit prime number $p>7$,such $p^{12}+5039\times 5041$ the factor number is less than $120$

I think maybe use Fermat theorem? How can solve it? Thank you

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  • $\begingroup$ Yes,That's you mean,Thank you $\endgroup$ – user94270 Nov 11 '13 at 7:35
  • $\begingroup$ Actually, do you consider only prime factors? $\endgroup$ – sve Nov 11 '13 at 7:38
  • $\begingroup$ No,is not only prime factors,Thank you $\endgroup$ – user94270 Nov 11 '13 at 7:45
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    $\begingroup$ I don't understand the question. Are you saying there is no prime $p\gt7$ such that that expression has a factor less than 120? or that there is no prime $p\gt7$ such that that expression doesn't have a factor less than 120? If $p\ge3$ is prime then the number is even, so it has 2 as a factor. The question is very unclear to me. $\endgroup$ – Gerry Myerson Nov 11 '13 at 9:23
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    $\begingroup$ @math110, but that's trivial, as for each prime exceeding 2 the expression has the factor 2. I think what's meant is that for every prime greater than 7 the expression has 120 as a factor, but that's sure not what it says. $\endgroup$ – Gerry Myerson Nov 11 '13 at 11:58
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Hint : $5040=7!$ , $120=5!$ , and $p=2k+1$. Our expression becomes: $$p^{12}+5039\cdot5041=p^{12}+(7!-1)(7!+1)=p^{12}+(7!^2-1)=(p^{12}-1)+7!^2=\\=(p^6-1)(p^6+1)+7!^2=(p^3-1)(p^3+1)\cdot(p^2+1)(p^4-p^2+1)+7!^2=\\=(p-1)(p^2+p+1)\cdot(p+1)(p^2-p+1)\cdot(p^2+1)(p^4-p^2+1)+5!^2\cdot42^2$$ $p>7>2\iff p=2k+1\iff(p-1),(p+1)$, and $p^2+1$ are all multiples of $2\iff$ the product is a multiple of $2^3=8$. Also, since $p>7>3\iff p=3k\pm1\iff p\mp1$ is a multiple of $3\iff$ the product is also a multiple of $3$. $p>7>5\iff p=5k\pm1$ or $p=5k\pm2$. In the former case, $p\mp1$ is a multiple of $5$, in the latter, $p^2+1$ is divisible through $5\iff$ the product is a multiple of $5$ as well, meaning that the entire sum divides through $5!=120$ , for all primes $p>7$


If however the total number of factors is what's actually meant, then let us first notice that the expression divides through $2$ once more, because $(p-1)(p+1)=2k(2k+2)=4k(k+1)$, and $k(k+1)$ is always even, generating an extra factor of $2$ ; through $3$ twice, because of $p^2\pm p+1$ ; as well as through $7$. $[\ p=7k\pm1\iff7|p\mp1$ ; $p=7k\pm2\iff7|p^2\pm p+1$ ; $p=7k\pm3$ $\iff7|p^2\mp p+1\ ]$. Thus, $N=2^4\cdot3^2\cdot5^1\cdot7^1\cdot n^1$ , implying a minimum of $(4+1)\cdot(2+1)\cdot$ $(1+1)\cdot(1+1)\cdot(1+1)=120$ factors. — Thanks to user Achille Hui for his invaluable help !

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  • $\begingroup$ then? Thank you, can you post your full solution? $\endgroup$ – china math Nov 11 '13 at 7:16
  • $\begingroup$ what was the question supposed to say? $\endgroup$ – Zackkenyon Nov 11 '13 at 7:17
  • $\begingroup$ Hello,why isn't exsit $p$ ?can you explain?,Thank you $\endgroup$ – user94270 Nov 11 '13 at 7:48
  • $\begingroup$ $5040 = 2^4 \cdot 3^2 \cdot 5 \cdot 7 | p^{12} - 1$ for prime $p > 7$. $\endgroup$ – achille hui Nov 11 '13 at 13:14
  • $\begingroup$ Yes, because $(p-1)(p+1)=2k(2k+2)=4k(k+1)$ , and $k(k+1)$ spits out yet another factor of $2$, thus increasing the total number of divisors to $60$. But how do we double it to $120$ ? $\endgroup$ – Lucian Nov 11 '13 at 13:39

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