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