Prove or disprove :

There exist 7 consecutive odd integers that are each divisible by a perfect cube greater than 1.

it is exist statement so if it is true we have to provide example and if it is false then we have to prove it

I see it is false statement because I do not find sequence like 7 consecutive odd integers that are each divisible by a perfect cube greater than 1.

let the 7 consecutive odd integers look like $2k+1 , 2k+3 , 2k+5,2k+7 ,2k+9 , 2k+11 $, and $2k+13$ are odd so if we select any positive integer $k$ there is at least one of odd integer will be prime

so , we cannot have cubic divisor

is my work correct or any suggestion on that? thanks

  • 2
    $\begingroup$ Have you heard of the Chinese Remainder Theorem? $\endgroup$ – Dylan Dec 11 '17 at 21:28
  • 5
    $\begingroup$ There is no reason why one of seven consecutive odd integers should be prime, so no, your work is not correct. $\endgroup$ – Malcolm Dec 11 '17 at 21:29
  • $\begingroup$ @Dylan If the penny doesn't drop soon, you might as well post that as an answer. It is easy to foresee that otherwise somebody else will :-/ $\endgroup$ – Jyrki Lahtonen Dec 11 '17 at 21:31
  • $\begingroup$ @JyrkiLahtonen , what would be wrong with that? $\endgroup$ – G Tony Jacobs Dec 11 '17 at 21:34
  • $\begingroup$ @Malcolm , what is the correct way that I can use ? thank you $\endgroup$ – dr.rise Dec 11 '17 at 22:04

If I wanted to prove this for $3$ consecutive odd integers, rather than $7,$ I would pick $3$ primes, say $5$, $7$, and $11$. Then I would use Chinese Remainder Theorem to solve the system

$$2n+1 \equiv 0 \pmod{ 5^3}$$ $$2n+3 \equiv 0 \pmod{7^3}$$ $$2n+5 \equiv 0 \pmod{11^3}$$

to get $n = 17242437$. The first of the $3$ integers is $34484875.$

  • $\begingroup$ Thank you very much, Do you have another way to solve it with out using Chinese Remainder Theorem ? $\endgroup$ – dr.rise Dec 11 '17 at 22:15
  • $\begingroup$ Not really. You can avoid CRT, but you'll have to do essentially the same calculations. $\endgroup$ – B. Goddard Dec 11 '17 at 22:23
  • $\begingroup$ $$2n+1 \equiv 0 \pmod{ 3^7}$$ $$2n+3 \equiv 0 \pmod{5^7}$$ $$2n+5 \equiv 0 \pmod{7^7}$$ $$2n+7 \equiv 0 \pmod{ 9^7}$$ $$2n+9 \equiv 0 \pmod{11^7}$$ $$2n+11 \equiv 0 \pmod{13^7}$$ $$2n+13 \equiv 0 \pmod{1^7}$$ so it will be look like that and after that I have to find n $\endgroup$ – dr.rise Dec 11 '17 at 22:25
  • 1
    $\begingroup$ Not quite. The moduli should be cubed. Also $9$ is not prime. Also the last modules should be $17^3$. $\endgroup$ – B. Goddard Dec 11 '17 at 22:30
  • $\begingroup$ $$2n+1 \equiv 0 \pmod{ 5^3}$$ $$2n+3 \equiv 0 \pmod{7^3}$$ $$2n+5 \equiv 0 \pmod{3^3}$$ $$2n+7 \equiv 0 \pmod{ 5^3}$$ $$2n+11 \equiv 0 \pmod{17^3}$$ $$2n+13 \equiv 0 \pmod{13^3}$$ $\endgroup$ – dr.rise Dec 15 '17 at 15:23

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