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I have a question which belongs to the field of number theory. Can we prove or disprove the following claim:

For all prime number $p=24t+1$ and the natural number $n=6t+1$, there is at least, one positive integer number $k$, such that the expression $\frac{k(n+1)}{7k-(n+1)}$ is a integer number.

I checked the above claim for prime numbers up to $17\times 10^9$, and the calculations verified it.

I will be so thankful for any helpful comments and answers.

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    $\begingroup$ Just a comment: write $m=n+1$; then your conditions tell you that $4m-7$ is prime and that $m \equiv 2 \pmod{6}$. In particular, $m$ is not divisible by $7$. Hence the expression $\frac{km}{7k-m}$ is an integer if and only if $d := 7k-m \mid m^2$. In particular, if $m \equiv 3$, $5$ or $6$ mod $7$, then we can find such a $k$, namely by setting $d = 4$, $2$ or $1$, respectively. $\endgroup$ – Tom De Medts Dec 22 '15 at 14:45
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    $\begingroup$ @individ: When $n=7=6\cdot 1 + 1$, the given expression is never a positive integer. In this case, however, $24 \cdot 1 + 1 = 25$ is not prime. So this indicates that at least some restriction on the number $n$ is necessary (although it could of course be that the given condition is stronger than what is needed). $\endgroup$ – Tom De Medts Dec 22 '15 at 14:52
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    $\begingroup$ In view of @Tom's remark, this is hardly possible if $m/2=3t+1$ is a prime congruent to 1, 2 or 4 modulo 7: then $m^2$ does not have any positive divisor congruent to $-m$ modulo 7. And I would say there SHOULD be such $t$. $\endgroup$ – Ilya Bogdanov Dec 22 '15 at 15:03
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    $\begingroup$ A counterexample: $t=14$, $p=337$, $n=85$. Thus $7k-86$ should divide $4\cdot 43^2$ which is impossible modulo 7. $\endgroup$ – Ilya Bogdanov Dec 22 '15 at 15:12
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    $\begingroup$ Meysam, it WAS a counterexample to your claim, since you asked the ratio to be a positive integer. Now the claim is true, since one may apply the same scheme with taking $-4,-2,-1$ into account. But I wouldnt say that changing the question without any notification in such situation is a good way of doing things. $\endgroup$ – Ilya Bogdanov Dec 22 '15 at 16:22

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