Show that ${(F_n^2+F_{n+1}^2+F_{n+2}^2)^2\over F_{n}^4+F_{n+1}^4+F_{n+2}^4}=2$

Question

If $F_n$ is the $n$-th Fibonacci number ($1,1,2,3,5,8,\dots$), show that

$${(F_n^2+F_{n+1}^2+F_{n+2}^2)^2\over F_{n}^4+F_{n+1}^4+F_{n+2}^4}=2$$

I have tested with a lot of Fibonacci numbers and it seem to obey the ruse, but I don't know how simplify it to 2.

I try:

Let $a=F_n$, $b=F_{n+1}$ and $c=F_{n+2}$

$a^4+b^4+c^4+2(ab)^2+2(ac)^2+2(bc)^2=2a^4+2b^4+2c^4$

$2(ab)^2+2(ac)^2+2(bc)^2=a^4+b^4+c^4$

I am not sure, what to do next. Can anyone help by completing the prove?

Answer 1

replace $c$ with $a+b$ and it should simplify.

$a^4+b^4+c^4 = a^4+b^4+(a+b)^4 = 2a^4+4a^3b+6a^2b^2+4ab^3+2b^4$

$2(ab)^2+2(ac)^2+2(bc)^2 = 2a^2b^2 + 2a^2(a^2+2ab+b^2) + 2b^2(a^2+2ab+b^2) \\ = 2a^4+4a^3b+6a^2b^2+4ab^3+2b^4$

Answer 2

As $F_{n+2}=F_{n+1}+F_n$. One first solves the characteristic polynomial. $$ X^2=X+1 $$
on gets two roots $r=\frac{1+\sqrt{5}}{2}$ and $\bar{r}=\frac{1-\sqrt{5}}{2}$. One now solves $\alpha.r^n+\beta.\bar{r}^n=F_n$ for the two first indices.

The OP did not indicate his initialization so I suppose that $F_0=0;\ F_1=1$, one gets then $F_n=\frac{1}{\sqrt{5}}(r^n-\bar{r}^n)$. This, with the relations $r+\bar{r}=1;\ r.\bar{r}=-1;\ r-\bar{r}=\sqrt{5}$ will, I think, be sufficient to get the desired result.

Coda In fact, it is true in general that $$ \Big(X^2+Y^2+(X+Y)^2\Big)^2=2\Big(X^4+Y^4+(X+Y)^4\Big) $$ thus, the fraction holds true for any sequence s.t. $s_{n+2}=s_n+s_{n+1}$.

Answer 3

Setting $x=a$, $y=b$, and $z=-(a+b)$ we have to evaluate $\frac{(x^2+y^2+z^2)^2}{x^4+y^4+z^4}$

Since $x+y+z= 0$, $x,y,z$ are roots of $t^3 + t(\sum xy)-xyz = 0$

Hence $x,y,z$ satisfy $t^4 + t^2(\sum xy)-xyzt = 0$

Setting $t=x,y,z$ successively and adding the resulting equations we obtain

$\sum x^4 +\sum x^2 \sum xy -xyz \sum x = 0 \implies \sum x^4 +\sum x^2 \sum xy = 0$

Since $\sum x = 0$ we get that $2\sum xy = - \sum x^2$

Now its easy to see that $\frac{(x^2+y^2+z^2)^2}{x^4+y^4+z^4} = 2$