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Find all positive integers $n$ such that the fraction $\dfrac{n+7}{2n+7}$ is the square of a rational number.

The answer says $n = 9,57,477$, but how do we prove this? I wrote $\dfrac{n+7}{2n+7} = \dfrac{a^2}{b^2}$ for some $a,b \in \mathbb{Z^+}$ and $\gcd(a,b) = 1$. We know that $\gcd(n+7,2n+7) = \gcd(n+7,7) = 1,7$ and so $n+7$ and $2n+7$ must both be perfect squares or their greatest common divisor is $7$. How do we find all values for which this is the case?

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    $\sqrt{10/19} \not \in \mathbb{Q}$ so how is $n=9$ a solution? – 2017-01-04
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    @dxiv Sorry, the $1$ should be a $7$. – 2017-01-04
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    You asked for a *proof* but you accepted an answer (really a comment) that is *far* from giving a proof. If you truly desire a proof then you should unaccept the answer (else many readers will skip the question, assuming it is answered) – 2017-01-04
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    The answers can likely be parameterized by the solutions to [Pell's equation](https://en.wikipedia.org/wiki/Pell's_equation). – 2017-01-04
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    Why is $\gcd(n+7,2n+7) = \gcd(n+1,7)$? Doesn't Euclid's algorithm give $\gcd(n+7,2n+7)=\gcd(n+7,n)=\gcd(7,n)$? – 2017-01-04
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    There cannot be any proof as the list of solutions is not complete. For the case studies one can just check that manually no further justification is required. It would be much less trivial to prove that this was a complete list of solutions, but that is clearly not the case, so the question is answered to the extend it make sense. – 2017-01-04
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    [DUPLICATE] of your recent question (http://math.stackexchange.com/q/2083696) which has received an answer – 2017-01-04

3 Answers 3

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$$\gcd(n+7,2n+7)=\gcd(n+7,n)=\gcd(7,n)\in\{1,7\}$$

If the $\gcd=1$, then $n+7,2n+7$ are both squares:

$$n+7=a^2,2n+7=b^2\implies 2a^2-b^2=7$$

If the $\gcd=7$, then $\frac{n+7}{7},\frac{2n+7}{7}$ are both integer squares. Write $n=7m$:

$$m+1=a^2,2m+1=b^2\implies2a^2-b^2=1$$

Finding exact solutions from here requires study of Pell's equation, as indicated in the comments.

  • The first equation has solutions starting with $\{(2,1),(8,11),\cdots\}$

  • The second equation has solutions starting with $\{(1,1),(5,7),(41,29),\cdots\}$.

These can be generated by looking at $(3+\sqrt2)(1+\sqrt2)^{2k},(1+\sqrt2)^{2k+1}$ respectively for integer $k$. Noting that $\frac{1}{1+\sqrt2}=-(1-\sqrt2)$ tells us that for the second of these sequences, we cann afford to simply look at $k\ge0$.

In the first case, given $2a^2-b^2=7$, note that taking $n=b^2-a^2$ gives $a^2=n+7,b^2=2n+7$ as desired, i.e. the fraction in question is a square. A slight adaptation of this resolves the second case.

Clarification: When I say "these can be generated by looking at $(a+b\sqrt2)^k$", I refer to the fact that by writing $(a+b\sqrt2)^k=A_k+B_k\sqrt2$ for integers $A_k,B_k$, we will have $2B_k^2-A_k^2=1$ or $7$, depending on this sequence. This method makes use of techniques from algebraic number theory, where we learn that the function $N:a+b\sqrt2\to a^2-2b^2$ is multiplicative.

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    Perhaps I am missing something, but I don't think this covers the case handled in $(3)$ in my answer. – 2017-01-05
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    Hi @robjohn, will have a look - my gut says that it may correspond to taking $k$ negative but I'll check. – 2017-01-05
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    As mentioned in my answer, it does match negative $k$, but there is no mention of negative $k$ or why we don't need to consider negative $k$ in the other list of solutions. – 2017-01-05
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    @robjohn True - will add in a comment reflecting this. I was more focused on getting to the Pell's equation than solving it for this question, so glossed over this side of things somewhat. – 2017-01-05
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Because $2(n+7)-(2n+7)=7$, we know that $\gcd(n+7,2n+7)\mid7$.

Since we only want positive $n$, this implies one of

  1. $n+7$ and $2n+7$ are squares, or

  2. $n+7$ and $2n+7$ are each $7$ times a square.

Case 1: Let $p^2=n+7$. Then we need $q^2=2n-7=2p^2-7$. This gives the Pell equation $$ 2p^2-q^2=7\tag{1} $$ Equation $(1)$ has two sequences of solutions: $$ (p,q)\in\{\overbrace{\ \ \ (2,1)\ \ \ }^{n=-3},\overbrace{\ \ (8,11)\ \ }^{n=57},\overbrace{\ (46,65)\ }^{n=2109},\dots\}\tag{2} $$ and $$ (p,q)\in\{\overbrace{\ \ \ (4,5)\ \ \ }^{n=9},\overbrace{\ (22,31)\ }^{n=477},\overbrace{(128,181)}^{n=16377},\dots\}\tag{3} $$ Both of which satisfy the recursion $$ (p_k,q_k)=6(p_{k-1},q_{k-1})-(p_{k-2},q_{k-2})\tag{4} $$ We can solve the recursion $(4)$ for $(2)$ with $$ p_k=\tfrac14\left(4+\sqrt2\right)\left(3+2\sqrt2\right)^k+\tfrac14\left(4-\sqrt2\right)\left(3-2\sqrt2\right)^k\tag{5} $$ and $$ q_k=\tfrac12\left(1+2\sqrt2\right)\left(3+2\sqrt2\right)^k+\tfrac12\left(1-2\sqrt2\right)\left(3-2\sqrt2\right)^k\tag{6} $$ Additionally, $(5)$ and $(6)$ with negative $k$ gives $(3)$ with negative $q$.

Therefore, we get the solutions for $n$ as $$ \begin{align} n_k &=p_k^2-7\\ &=\bbox[5px,border:2px solid #C0A000]{\frac18\left[\left(9+4\sqrt2\right)\left(3+2\sqrt2\right)^{2k}+\left(9-4\sqrt2\right)\left(3-2\sqrt2\right)^{2k}-42\right]}\tag{7} \end{align} $$ where $k\in\mathbb{Z}$. For example, if $k\in\{-5,-4,-3,-2,-1,0,1,2,3,4,5\}$, $$ n_k\in\{18905097,556509,16377,477,9,-3,57,2109,71817,2439837,82882809\} $$

Case 2: Let $7p^2=n+7$. Then we need $7q^2=2n+7=14p^2-7$. As above, this leads to the Pell's Equation $$ 2p^2-q^2=1\tag{8} $$ Equation $(8)$ has the solutions $$ (p,q)\in\{\overbrace{\ \ \ (1,1)\ \ \ }^{n=0},\overbrace{\ \ \ (5,7)\ \ \ }^{n=168},\overbrace{\ (29,41)\ }^{n=5880},\dots\}\tag{9} $$ which satisfies $(4)$.

We can solve the recursion $(4)$ for $(9)$ with $$ p_k=\tfrac14\left(2+\sqrt2\right)\left(3+2\sqrt2\right)^k+\tfrac14\left(2-\sqrt2\right)\left(3-2\sqrt2\right)^k\tag{10} $$ and $$ q_k=\tfrac14\left(2+2\sqrt2\right)\left(3+2\sqrt2\right)^k+\tfrac14\left(2-2\sqrt2\right)\left(3-2\sqrt2\right)^k\tag{11} $$ Formulas $(10)$ and $(11)$ with negaitve $k$ just give $(9)$ with negative $q$. Thus, we only need consider $k\ge0$. $$ \begin{align} n_k &=7p_k^2-7\\ &=\bbox[5px,border:2px solid #C0A000]{\frac78\left[\left(3+2\sqrt2\right)^{2k+1}+\left(3-2\sqrt2\right)^{2k+1}-6\right]}\tag{12} \end{align} $$ where $k\in\mathbb{Z}^{\ge0}$. For exmple, if $k\in\{0,1,2,3,4,5\}$, $$ n_k\in\{0,168,5880,199920,6791568,230713560\} $$

Complete Closed and Recursive Solutions

Collecting the results in $(7)$ and $(12)$, and reindexing for monotonicity with $k\ge1$, we get that the closed form for the complete list of solutions is $$ \hspace{-8mm}\bbox[5px,border:2px solid #C0A000]{ \begin{bmatrix}n_{3k-2}\\n_{3k-1}\\n_{3k}\end{bmatrix} =\frac18\left(\,\begin{bmatrix}9-4\sqrt2\\9+4\sqrt2\\21+14\sqrt2\end{bmatrix}\left(3+2\sqrt2\right)^{2k}+\begin{bmatrix}9+4\sqrt2\\9-4\sqrt2\\21-14\sqrt2\end{bmatrix}\left(3-2\sqrt2\right)^{2k}-\begin{bmatrix}42\\42\\42\end{bmatrix}\,\right)}\tag{13} $$ and a recursive list of solutions is given by
$\scriptsize\begin{bmatrix}n_1\\n_2\\n_3\end{bmatrix}=\begin{bmatrix}9\\57\\168\end{bmatrix},\quad\scriptsize\begin{bmatrix}n_4\\n_5\\n_6\end{bmatrix}=\begin{bmatrix}477\\2109\\5880\end{bmatrix},\quad\scriptsize\begin{bmatrix}n_7\\n_8\\n_9\end{bmatrix}=\begin{bmatrix}16377\\71817\\199920\end{bmatrix},\quad$ and $$ \bbox[5px,border:2px solid #C0A000]{ \begin{bmatrix}n_{3k+1}\\n_{3k+2}\\n_{3k+3}\end{bmatrix}=35\begin{bmatrix}n_{3k-2}\\n_{3k-1}\\n_{3k}\end{bmatrix}-35\begin{bmatrix}n_{3k-5}\\n_{3k-4}\\n_{3k-3}\end{bmatrix}+\begin{bmatrix}n_{3k-8}\\n_{3k-7}\\n_{3k-6}\end{bmatrix} }\tag{14} $$

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There are more solutions (n<1000000): $$ \begin{array}{c} 9 \\ 57 \\ 168 \\ 477 \\ 2109 \\ 5880 \\ 16377 \\ 71817 \\ 199920 \\ 556509 \\ \end{array} $$ so you just plug these values of $n$ and check otherwise the question does not make sense. Also One solution in your list is missing for n<500.

Added after observation of ross-millikan:

$n+7$ and $2n+7$ can have common multiplier, is so than $2*(n+7)-(2n+7)$ is also devisible by the same number which is clearly $7$, so the theorem is that if numerator and denominator have a common multiplier it can only be $7$. Thus we have two possibilities

a) $n+7=a^2$ and $2n+7=b^2$

b) $n+7=7 a^2$ and $2n+7=7 b^2$

for b) n should be divisible by $7$

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    This looks like two families of solutions to a Pell-like equation. $9,57,477,2109$ and others lead to $n+7,2n+7$ both being squares. Every third one, $168, 5880, \ldots$ do not. The second batch have $n+7, 2n+7$ sharing a factor $7$ – 2017-01-04
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    good point. so if n+7 and 2n+7 have common factor then so much have 2(n+7)-(2n+7)=7 and it has only 1 and 7 as common factors so that explains two families of solutions! – 2017-01-04