Everything that follows takes place in $\mathbb F_5(\sqrt 2)$, i.e. the field we get by adjoining a root of $x^2-2=0$ to $\mathbb F_5$, the field with $5$ elements. We have $\sum_{k=0}^n\binom{2n+1}{2k+1}2^{3k}=\sum_{k=0}^n\binom{2n+1}{2n-2k}3^k=\sum_{k=0}^n\binom{2n+1}{2(n-k)}2^{-k}$. Now, this is zero iff it's zero when we multiply it by $2^n$, so we may as well prove that $\sum_{k=0}^n\binom{2n+1}{2(n-k)}\sqrt 2^{2(n-k)}\ne 0$. The LHS is $\alpha$ from $(1+\sqrt 2)^{2n+1}=\alpha+\beta\sqrt 2,\ \alpha,\beta\in\mathbb F_5$. We have $(1-\sqrt 2)^{2n+1}=\alpha-\beta\sqrt 2$, so by multiplying them we get $-1=\alpha^2-2\beta^2$. If we were to have $\alpha=0$, then we would get $1=2\beta^2,\ \beta\in\mathbb F_5$, and this is impossible, since it would make $3=2^{-1}$ a square $\beta^2$ in $\mathbb F_5$ (i.e. $3$ would be a quadratic residue modulo $5$, and it's not).

how do you make the pass from $2^{3k}$ to $3^k$ and then to $2^{-k}$?

He works $\mod 5$ I think... (or better to say: in it's quadratic extension $\mathbb{F}_{25}$)

he says he works in $\mathbb F_5 (\sqrt{2})$. how is that the same thing?

$\mathbb{F}_5$ is just a synonym for the field of integers $\mod 5$. There is no solution $a^2 \equiv 2 \mod 5$, so adjoin $\sqrt 2$, which means just consider all numbers of the form $a+b\sqrt 2 \mod 5$ with $a,b$ residue classes $\mod 5$ (all together written as $\mathbb{F}_5[\sqrt 2]$). There are $25$ of them and they build up a field, also called $\mathbb{F}_{25}$.

Thanks. I finally understood the solution.

i have a simpler (though a little longer) solution: re-write your sum: \[ a_n = \sum_{k = 0}^{n} \binom{2n + 1}{2k + 1} 2^{3k} = \frac {(1 + 2\sqrt2)^{2n + 1} + (1 - 2\sqrt2)^{2n + 1}}{2} \] now, what i've wrote is a sequence, generated by recurrence: $ a_{n + 1} = 18a_n - 49a_{n - 1}$, whose first terms are $ a_0 = 1, a_1 = 11$. it's an easy check to show that no terms can be $ 0 \mod 5$, since sequence has a short period...

ma_go wrote: now, what i've wrote is a sequence, generated by recurrence: $ a_{n + 1} = 18a_n + 49a_{n - 1}$ That should be $ a_{n+1}=18a_n-49a_{n-1}$ and it will work indeed Pretty nice solution anyways.

$$ \sum\binom{2n+1}{2k+1} 2^{3k}= \sum\binom{2n+1}{2k+1} 3^{k} (mod 5)$$$$(1+3)^{2n+1}= \sum\binom{2n+1}{r} 3^{r}$$$$ \sum \binom{2n+1}{2k+1} 3^{k}=4^{2n+1}-\left(\sum \binom{2n+1}{2k} (-1)^{k}\right)$$$$(1+i)^{2n+1}= \sum\binom{2n+1}{2} i^{k}$$$$4^{2n+1}-\left(\sum \binom{2n+1}{2k} (-1)^{k}\right)=4^{2n+1}-Re((1+i)^{2n+1})=4^{2n+1}-2^n \sqrt{2} \cos\left((2n+1) \frac{\pi}{4}\right)$$ The rest is just checking mod 5. Is my solution correct?

I don't fully understand about group, field, etc (high algebra), but I know binomial expansion: $(1\pm x)^m=\sum_{k=0}^{m}{\dbinom{m}{k}(\pm x)^k}=\sum_{k=0}^{m}{\dbinom{m}{2k}x^{2k}}\pm\sum_{k=0}^{m}{\dbinom{m}{2k+1}x^{2k+1}}=\sum_{k=0}^{m}{\dbinom{m}{2k}(x^2)^k}\pm x\sum_{k=0}^{m}{\dbinom{m}{2k+1}(x^2)^k}$ So if we set $m=2n+1$ and $x^2=2^3 \iff x=\sqrt{2^3}$, we get: $(1\pm \sqrt{2^3})^{2n+1}=\sum_{k=0}^{2n+1}{\dbinom{2n+1}{2k}(2^3)^k}\pm \sqrt{2^3}\sum_{k=0}^{2n+1}{\dbinom{2n+1}{2k+1}(2^3)^k}=\sum_{k=0}^{2n+1}{\dbinom{2n+1}{2k}2^{3k}}\pm \sqrt{2^3}\sum_{k=0}^{2n+1}{\dbinom{2n+1}{2k+1}2^{3k}}=a\pm b\sqrt{2^3}$ We have to show $b=\sum_{k=0}^{2n+1}{\dbinom{2n+1}{2k+1}2^{3k}}$ is not a multiply of 5. $(a+b\sqrt{2^3})(a-b\sqrt{2^3})=(1+\sqrt{2^3})^{2n+1}\cdot (1-\sqrt{2^3})^{2n+1}$ $\iff a^2-8b^2=[(1+\sqrt{2^3})(1-\sqrt{2^3})]^{2n+1} $ $\iff a^2-8b^2=(1-(\sqrt{2^3})^2)^{2n+1}=(-7)^{2n+1}=-7^{2n+1}$ $\iff 8b^2=a^2+7^{2n+1}$ Because: * $a\equiv 0 (mod 5) \to a^2 \equiv 0 (mod 5)$ * $a\equiv \pm 1 (mod 5) \to a^2 \equiv 1 (mod 5)$ * $a\equiv \pm 2 (mod 5) \to a^2 \equiv 4 \equiv -1 (mod 5)$ so: $7^{2n+1}=7\cdot (7^n)^2 \equiv 2\cdot \pm 1 \equiv \pm 2 (mod 5)$ It follows: $8b^2=a^2+7^{2n+1} \not\equiv 0 (mod 5) \to b \not\equiv 0 (mod 5)$ q.e.d