How many positive integers $n$ are there such that $(7n + 1)$ is a perfect square and $(3n + 1) < 2008$?

Question

How many positive integers $n$ are there such that $(7n + 1)$ is a perfect square and $(3n + 1) < 2008$ ?

What I Tried: We have $(3n + 1) < 2008$ $\rightarrow n < 669$ .
Now, $(7n + 1) = k^2$ for some $k$ . $$\rightarrow 7n = (k + 1)(k - 1)$$ Now I have to count all the $n$ so that after multiplied by $7$ , gives a product same as the product of $2$ consecutive numbers with a gap of one. I know that $n = 5,9$ are two solutions, but there are more values so that when multiplied by $7$ gives a valid solution, like $k = 24,32$ .

The problem I am facing is, is there any elegant way here to count all the solutions?
Can anyone help me?

Answer 1

As you mentioned, $7n = (k + 1)(k - 1)$

So $7$ is a factor of either $(k+1)$ or $(k-1)$

which means $k = 7 a -1$ or $7a+1$.

If $k = 7a - 1$, $n = 7a^2 - 2a \lt 669$ as you observed.

Taking $a = 1, 2, ...$ you get the values but if you only have to count the number of solutions, it is easy to see that $a \leq 9$ given the constraint $n \lt 669$.

Similarly take $k = 7a + 1$ and see how many solutions given the constraint of $n \lt 669$.

Answer 2

Since $n\le 668$, we know that $7n+1\le 4677$. $\left\lfloor\sqrt{4677}\right\rfloor=68$, so we want to know how many of the integers $1,2,\ldots,68$ have squares of the form $7n+1$. Suppose that $7n+1=k^2$ for some integer $k$; then $k^2\equiv 1\pmod 7$, and it’s easy to check that this is the case precisely when $k\equiv 1\pmod 7$ or $k\equiv -1\pmod 7$, i.e., when $k=7m\pm 1$ for some integer $m$.

There are $9$ positive multiples of $7$ less than $68$, each of which provides two such values of $k$.

E.g., $3\cdot 7=21$ means that $k=20$ and $k=22$ will work: $20^2=400=7\cdot 57+1$, and $22^2=484=7\cdot 69+1$.

Thus, there are altogether $2\cdot9=18$ such integers $n$.

Answer 3

You just have to find the number of solutions.

$k$ such that $(k+1)$ or $(k-1)$ is divisible by $7$ and $k>1$.

Note that $7n+1 < 7\cdot 669+1=4684$
Therefore, $k<69$

Every $k$ such that $(k+1)$ or $(k-1)$ is divisible by $7$ and $69>k>1$ will give you a solution.

Therefore, $k = 6,13,20,...62$ These are $\frac{62-6}{7}+1 = 9$ numbers

or $k = 8,15,22,...,64$ These are $\frac{64-8}{7}+1 = 9$ numbers

$9+9=18$ solutions