show that $a_{n}=\frac{1}{4}[(1+\sqrt{2})^{2n+1}+(1-\sqrt{2})^{2n+1}+2](n>1)$ not have square numbers

Question

Show that the sequence $$a_{n}=\dfrac{1}{4}[(1+\sqrt{2})^{2n+1}+(1-\sqrt{2})^{2n+1}+2]\qquad (n>1)$$ doesn't contain a perfect square.

I think this problem is interesting, and my idea: we can find $A$, $B$ and $C$ constants $$a_{n+1}=Aa_{n}+Ba_{n-1}+C$$

Thank you

Did you mean to show $,a_n,$ is an integer (or even a natural) number which isn't divisible by the square of a natural number? Say, not like $,12,$ which is divisible by $,2^2=4,$ , etc.? — DonAntonio, Jun 08 '13 at 10:35
I mean $a_{n}\neq m^2,m\in N^{+}$,for all $n\ge 2$,Thank you — math110, Jun 08 '13 at 10:37
The hint that was given to you is probably rather to fint $A$, $B$ and $C$ such that $a_{n+2}=Aa_{n+1}+Ba_n+C$ for every $n\geqslant0$. Can you confirm this is the hint and show what you did to apply it? — Did, Jun 08 '13 at 10:49
@Did, these WHYT comments (and the ideas/tried/work type of information volunteered by the OP) are rather pointless, because playing with the recurrence goes nowhere for this type of problem. That is in fact an essential piece of information about the problem. Either you don't know how to answer the question (in which case, why comment so much?), or you do and are leading the OP in circles in the name of What Have You Tried. It's pretty obvious this poster is a problem-solving enthusiast and some of the problems posted have been fairly deep, so his sources or taste in material are good. — zyx, Jun 10 '13 at 17:40

score 7 · Answer 1 · edited Apr 13 '17 at 12:19

$a_n$ are coordinates ($x$, let's say) of integer points on a hyperbola with an equation $F(x,y)=0$. The number of integer points on $F(x^2,y)=0$ is finite, as this equation is of a genus $1$ curve, and the points can be found by known methods. The proof is either something trivial using congruences to show no solutions exist (usually not possible, this is an easy but less common case), or an analysis of Thue equations (usually done by software these days, and can involve linear forms in logarithms or other machinery as part of the algorithm).

Because the software exists, normally people do not look for elementary proofs of these things "by bare hands" without a good reason. If this is a contest problem there is a proof using congruences or a nice factorization of the equation, otherwise expect to use a computer program or to spend time learning about Thue equations (which will convince you to use a computer program).

Here is one way to get the answer. To find $F(x,y)$, introduce a sequence $b_n = C((1 + \sqrt{2})^n - (1 - \sqrt{2})^n)$ with $C$ chosen to obtain a linear relation between $(4a_n - 2)^2$ and $b_{2n+1}^2$. In coordinates $x=a_n$ and $y=b_{2n+1}$, this relation is a degree $2$ equation $F(x,y)=0$. Then convert $F(x^2,y)=0$ to Weierstrass form and input to SAGE as explained in links here: How to compute rational or integer points on elliptic curves . If there are solutions, check if they are in the sequence $a_n$.

This might be a little off topic here, but how would one find all the integral points for the curve we have in hand? When I follow your link, what I would do is to first transform the curve into Weierstrass form and find the integer points on it. But I suppose the transformation of the curve to Weierstrass form is defined over $\mathbb{Q}$ not $\mathbb{Z}$ so I'm not exactly finding the points I want. Is there a way to do it directly? — , Jun 10 '13 at 19:24
The program can find integral points over $Z[1/n]$ for given $n$. I'm not sure about S-integral points for finite S. @Sanchez — zyx, Jun 10 '13 at 19:34
sure. The thing is if I don't want to go through the process of finding the isomorphism to Weierstrass form so as to locate the denominator and then to transform the points in Weierstrass form back to the original curve etc, is it possible? Is there any known function in Magma/Sage that allows me to compute the integral points on any non-rational curves given in the form of plane curve? — , Jun 10 '13 at 19:41
I forgot to list descent as another possible elementary solution. — zyx, Jun 11 '13 at 22:31

score 5 · Accepted Answer · answered Jun 08 '13 at 21:43

This is not a solution, but it may be of some use. Your conjecture about a recurrence for the $a_n$ is correct.

Let $\alpha=1+\sqrt2$ and $\beta=1-\sqrt2$. Let $\gamma=\alpha^2=3+2\sqrt2$ and $\delta=\beta^2=3-2\sqrt2$. The recurrence can then be written

$$a_n=\frac\alpha4\gamma^n+\frac\beta4\delta^n+\frac12\;.$$

Let

$$b_n=a_n-\frac12=\frac\alpha4\gamma^n+\frac\beta4\delta^n\;;$$

$\gamma+\delta=6$ and $\gamma\delta=1$, so $\gamma$ and $\delta$ are the roots of the quadratic

$$0=(x-\gamma)(x-\delta)=x^2-(\gamma+\delta)x+\gamma\delta=x^2-6x+1\;,$$

so by the general theory of linear homogeneous recurrences we know that the $b_n$ satisfy the recurrence

$$b_n=6b_{n-1}-b_{n-2}\;.$$

This can be rewritten

$$a_n-\frac12=6\left(a_{n-1}-\frac12\right)-\left(a_{n-2}-\frac12\right)\;,$$

or

$$a_n=6a_{n-1}-a_{n-2}-2\;.$$

From the original definition we see that

$$a_0=\frac14(\alpha+\beta)+\frac12=1$$

and

$$a_1=\frac14\left(\alpha^3+\beta^3\right)+\frac12=4\;.$$

The sequence therefore begins $1,4,21,120,697,\dots\;$, and it turns out to be OEIS A182435. That entry in OEIS isn’t very helpful, but this sequence is also OEIS A046090 with the index offset by $1$; neither entry seems to mention the fact that they’re essentially the same sequence, but the second formula give for A046090 confirms it.

The header for A046090 reads as follows:

Consider all Pythagorean triples $(X,X+1,Z)$ ordered by increasing $Z$; sequence gives $X+1$ values.

This may be of help in showing that $a_n$ is not a square for $n\ge 2$; I’ve not had time to look at it very hard.

Thank you,It's good job,I think this problem hava other nice methods. — math110, Jun 10 '13 at 16:39
Thanks for locating the OEIS entries. Pythagorean triples means that the present Diophantine problem is $(X^2 - 1)^2 + (X^2)^2 = Z^2$. After putting into Weierstrass form, that equation can be fed to SAGE or Pari to get the integer points. — zyx, Jun 10 '13 at 18:11

score 2 · Answer 3 · answered Jun 13 '13 at 22:27

This is no complete solution, but may help bring $a_n$ somewhat incontext with "nearby" squares.

One readily checks that
$$a_0=1,\quad a_1=4, \quad a_n=6a_{n-1}-a_{n-2}+2.$$ Define the sequence $(b_n)_n$ by $$b_0 = 1, \quad b_1=1,\quad b_n=2b_{n-1}+b_{n-2}-1,$$ which amounts to $$b_n=\frac14\left((1+\sqrt2)^n+(1-\sqrt 2)^n+2\right) $$ Then $a_n=b_{2n+1}$. We observe that $b_{4n}$ and $b_{4n+2}-1$ are squares for all $n$. In fact, from $(1+\sqrt 2)(1-\sqrt 2)=-1$ we see that $$ \left(\frac12\left((1+\sqrt2)^{k}+(1-\sqrt 2)^{k}\right)\right)^2=\frac14\left((1+\sqrt2)^{2k}+(1-\sqrt2)^{2k}+2(-1)^k\right),$$ that is $b_{4m}=(2b_{2m}-1)^2$ and $b_{4m+2}=(2b_{2m+1}-1)^2+1$ Now if $n=2m$ is even, we obtain $$ b_{4m+2}=2b_{2n+1}+b_{4m}-1$$ i.e. $$ (2b_{2m+1}-1)^2+1=2a_n+(2b_{2m}-1)^2-1$$ or $$a_n-1 = 2(b_{2m+1}+b_{2m}-1)(b_{2m+1}-b_{2m}) = 2(b_{2m+2}-b_{2m+1})(b_{2m+1}-b_{2m})=2c_{2m+1}c_{2m},$$ where $$c_0=0,\quad c_1=1,\quad c_n=2c_{n-1}+c_{n-2}. $$ If $n$ is odd, we get a similar result with $a_n$ instead of $a_n-1$.

Minor: There are a few places where you change $m$ and $n$, like $b_{4m+2} = 2 {2n+1} + b{4m} -1$ and the equation after it. — Calvin Lin, Jun 14 '13 at 23:55

show that $a_{n}=\frac{1}{4}[(1+\sqrt{2})^{2n+1}+(1-\sqrt{2})^{2n+1}+2](n>1)$ not have square numbers

3 Answers3