11
$\begingroup$

I read about this problem back in secondary school and never came close to producing a proof. It has bugged me ever since. I suspect I'm not familiar enough with number theory tricks and so forth to solve it (as well as not being clever enough).

The question is, prove that when N is a whole number it is also a perfect square, where

$$N = 2 + 2\sqrt{12n^2 + 1}$$

so, for example, $N = 16$ when $n = 2$.

Cheers

P.S. this is the first time I have used this site so apologies in advance for defying standard conventions.

$\endgroup$
5
  • $\begingroup$ For community reference: the first few values of $n$ for which $N$ is an integer are $2,28,390,5432$. This list is exhaustive up to $10^6$ $\endgroup$ – Ben Grossmann Aug 6 '13 at 2:12
  • $\begingroup$ @Omnomnomnom, I found those values and also $n=75658$ (hopefully now exhaustive up to $10^6$). $\endgroup$ – Yoni Rozenshein Aug 6 '13 at 2:13
  • $\begingroup$ @YoniRozenshein Thanks, I just found that after noticing my program had crashed mid-computation. Now that it ran to completion, it seems the list is properly exhaustive. $\endgroup$ – Ben Grossmann Aug 6 '13 at 2:15
  • $\begingroup$ Observation (no proof, just from the above list): The largest prime factor of $n$ and $\sqrt N$ is the same. For the 5 cases up to $10^6$ this prime is $2, 7, 13, 97, 181$. $\endgroup$ – Yoni Rozenshein Aug 6 '13 at 2:15
  • $\begingroup$ @YoniRozenshein I added an explanation for your observation, which was interesting. $\endgroup$ – Calvin Lin Aug 6 '13 at 2:38
6
$\begingroup$

Hint: Pell's equation, since $ 12n^2 + 1 = m^2 $ allows you to classify the values.

Sorry but I don't know any other approach, if you are not familiar with this number theory trick.


Observe that in order of $N$ to be an integer, $12n^2 + 1$ must be the square of a rational number. Since it is an integer, hence it must be a perfect square. Thus we have the Pell's equation of the form $m^2 - 12n^2 = 1 $.

We can check that $7^2- 12 \times 2^2=1$ is an initial solution. From the theory of Pell's equation, we know that the solutions have the form $$m_i + \sqrt{12} n_i = (7 + 2\sqrt{12})^i = ( 2 + \sqrt{3})^{2i}.$$

Hence,

$$2 + 2 m_i = 2 + (2+\sqrt{3})^{2i} + (2-\sqrt{3})^{2i} = [(2 +\sqrt{3})^i + (2-\sqrt{3})^i]^2. $$

It is now obvious that the expression in the brackets is an integer, hence $2+2m_i$ is a perfect square.


To motivate Yoni's observation that "The largest prime factor of $n$ and $N$ are the same", observe that

$$n_i = \frac{ (2+\sqrt{3})^{2i} - (2-\sqrt{3})^{2i} } {\sqrt{12} } = N_i \frac{ (2+\sqrt{3})^{i} - (2-\sqrt{3})^{i}}{\sqrt{12} }$$

The second term on the right is an integer that is much smaller than $N_i$.

$\endgroup$
0
$\begingroup$

If 12n^2+1 is a square, m^2, then m^2-12n^2=1. This is Pell's equation; looking at the continued fraction expansion of $\sqrt{12}$ shows that 7^2-12*2^2 is the smallest solution. So all values of m, n are given by $m+n\sqrt{12} = (7+2\sqrt{12})^k$ for some positive integer k.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.