I am having difficulty understanding the solution. What is k? Please explain how is this solved using k :(
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$\begingroup$ We need to have $\frac q4$ be in integer. We don't know which integer, it could be any integer. So we call it $k $. $\endgroup$– fleabloodCommented Jan 25, 2018 at 3:57
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2 Answers
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Note that $p+1=7\cdot\frac{q}{4}$. Since $p$ is a natural number, and $4$ and $7$ are coprime, implies, $q$ must be divisible by $4$. By similar arguments, one can say that $p+1$ is divisible by $7$.
In fact, you get: $$\frac{p+1}{7}=\frac{q}{4}$$ which hints at the fact that the result of dividing $p+1$ by $7$, or dividing $q$ by $4$ will yield the same natural number (since they both are divisible). $k$ is that same natural number.
Once you arrive at $p+1=7k$ and $q=4k$, you can plug those values into the inequalities as the book has done. Since $k$ is a natural number, at the last step, you're able to narrow down the number of possible valid solutions from $k\leq93/7$ to simply $k=1,2,3,...,13$. This is the role of $k$.
answered Jan 25, 2018 at 3:51
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$\begingroup$ Instead of doing (p + 1)/7 = q/4 = k, why can't we do 4*(p + 1) / 7 = q = k ? $\endgroup$ Commented Jan 25, 2018 at 5:02
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1$\begingroup$ @arandomguy then you would get the value of $p=\frac{7k}{4}-1$. Now, you'll follow the same procedure as given in the book and get $k\leq372/7$ i.e., $k=1,2,3,...,53$ But, now note that, since $p$ is natural, and $p=\frac{7k}{4}-1$, so, $k$ has to be divisible by $4$. This means, here $k$ can only be $4,8,...,52$ i.e. 13 values. Hence, you can take $4\cdot\frac{p+1}{7}=q=k$, but it will result in one more extra step, and so it is not the best method. Hope it helps! $\endgroup$ Commented Jan 25, 2018 at 5:13
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As $q$ is an integer, $7$ must divide $4(p+1)$
Hence $7|(p+1)$ as $(4,7)=1$
$\implies p +1$ can be written as $7k$ where $k$ is any integer
answered Jan 25, 2018 at 3:50