2
$\begingroup$

Say a rational number is $kingly$ if it is the square of another rational number. Find the sum of all positive integers $n$ such that $\frac{n+7}{25n+7}$ is kingly.

If $\dfrac{n+7}{25n+7}$ is kingly, then we can write $\dfrac{n+7}{25n+7} = \dfrac{p^2}{q^2}$ for some $p,q \in \mathbb{Z}^+$ and $\gcd(p,q) = 1$. Rearranging the equation gives $$q^2(n+7) = p^2(25n+7).$$

How can I continue?

  • 1
    from what you have, either $n+7$ and $25n+7$ are both perfect squares (which is not possible). or $(n+7)$ divides $n+25$, and the ratio is a perfect square. The limit as n goes to infinity $\frac {25n+7}{n+7} = 25$ So, there are 3 possible values that $\frac {25n+7}{n+7}$ could take on where the ratio is a perfect square. i.e. check to see if there is an $n$ such that $\frac {25n+7}{n+7} = 4,9$ or $16$ – 2017-01-03
  • 0
    @DougM Is the answer $1$? – 2017-01-03
  • 0
    @DougM, could you expand your comment into an answer? – 2017-01-03
  • 1
    @DougM Why does $(n+7)$ have to divide $(n+25)$? Why can't they just share a few divisors. – 2017-01-03
  • 0
    Hmmm. I have found a mistake in my logic. $n=20$ is a solution. – 2017-01-03
  • 0
    "Find the sum of all positive integers $n$ such that..." sounds a lot like the format of a contest problem, because there is absolutely no motivation for only wanting the sum and not the entire set. I'm guessing there is a source for this problem that you have once again neglected to provide. – 2017-01-03

1 Answers 1

1

Option 1: $n+7$ and $25n+7$ are both perfect squares

Not possible:

All perfect squares are equivalent to $0,1$ or $4$ modulo $5$

$25n+7\equiv 2\pmod 5$

Option 2: $(n+7)$ divides $(n+25)$ and the ratio is a perfect square.

$\lim_\limits {n\to \infty} \frac {n+7}{25n+7}= \frac 1{25}$

There are 3 possible values that $\frac {n+7}{25n+7}$ could take on where the ratio is a perfect square. $\frac {n+7}{25n+7} = \frac 14 \implies n =1$

$\frac {n+7}{25n+7} = \frac 1{9} \implies 16 n = 56$ and $n$ is not an integer.

$\frac {n+7}{25n+7} = \frac 1{16} \implies 9 n = 105$ and $n$ is not an integer.

Indeed I have overlooked something.

$n = 20 \implies \frac {27}{507} = \frac {9}{169}$

$p^2(25n+7) = q^2(n+7)\\ (25p^2 - q^2) n = 7(q^2-p^2)$

$(5p+q) | 7(q+p)(q-p)$ and $5p-q |7(q+p)(q-p)$

  • 0
    Does this prove there are no others? – 2017-01-04
  • 0
    Nope. I found two answers and I was only expecting to find 1. – 2017-01-04