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The integral of each individual monomial will be integral. First we have the identity $$F_n(x)=\sum_{i=0}^{\lfloor(n-1)/2\rfloor}{n-i-1\choose i}x^{n-2i-1},$$ so my claim is that $$\binom{n-i-1}{i}\int_0^1 (k+nz)^{n-2i-1}dz=\binom{n-i-1}{i}\cdot\frac{(k+n)^{n-2i}-k^{n-2i}}{n(n-2i)} \in \mathbb Z.$$ By the binomial theorem we can write $(k+n)^{n-2i}=k^{n-2i}+nk^{n-2i-1}(n-2i)+n^2d$ for some integer $d$. So we can write $$\binom{n-i-1}{i}\cdot\frac{(k+n)^{n-2i}-k^{n-2i}}{n(n-2i)}=k^{n-2i-1}\binom{n-i-1}{i}+d\cdot\frac{n}{n-2i}\binom{n-i-1}{i}$$ it suffices to show that $\frac{n}{n-2i}\binom{n-i-1}{i}$ is an integer. However we can check that $\frac{n}{n-2i}\binom{n-i-1}{i}=\binom{n-i-1}{i}+2\binom{n-i-1}{i-1}$ and the claim follows.

The integral of each individual monomial will be integral. First we have the identity $$F_n(x)=\sum_{i=0}^{\lfloor(n-1)/2\rfloor}{n-i-1\choose i}x^{n-2i-1},$$ so my claim is that $$\binom{n-i-1}{i}\int_0^1 (k+nz)^{n-2i-1}dz=\binom{n-i-1}{i}\cdot\frac{(k+n)^{n-2i}-k^{n-2i}}{n(n-2i)} \in \mathbb Z.$$ By the binomial theorem we can write $(k+n)^{n-2i}=k^{n-2i}+nk^{n-2i-1}(n-2i)+n^2d$ for some integer $d$. So we can write $$\binom{n-i-1}{i}\cdot\frac{(k+n)^{n-2i}-k^{n-2i}}{n(n-2i)}=k^{n-2i-1}\binom{n-i-1}{i}+d\cdot\frac{n}{n-2i}\binom{n-i-1}{i}$$ it suffices to show that $\frac{n}{n-2i}\binom{n-i-1}{i}$ is an integer. However we can check that $$\frac{n}{n-2i}\binom{n-i-1}{i}=\binom{n-i-1}{i}+2\binom{n-i-1}{i-1}$$ and the claim follows.

The integral of each individual monomial will be integral. First we have the identity $$F_n(x)=\sum_{i=0}^{\lfloor(n-1)/2\rfloor}{n-i-1\choose i}x^{n-2i-1},$$ so my claim is that $$\binom{n-i-1}{i}\int_0^1 (k+nz)^{n-2i-1}dz \in \mathbb Z.$$ Since $\int_0^1 (k+nz)^{n-2i-1}dz=\frac{(k+n)^{n-2i}-k^{n-2i}}{n(n-2i)}$ and $(k+n)^{n-2i}-k^{n-2i}=nk^{n-2i-1}(n-2i)+n^2d$ for some integer $d$, we have $$\binom{n-i-1}{i}\cdot\frac{(k+n)^{n-2i}-k^{n-2i}}{n(n-2i)}=k^{n-2i-1}\binom{n-i-1}{i}+d\cdot\frac{n}{n-2i}\binom{n-i-1}{i}$$ it suffices to show that $$\frac{n}{n-2i}\binom{n-i-1}{i}\in \mathbb Z.$$ However $\frac{n}{n-2i}\binom{n-i-1}{i}=\binom{n-i-1}{i}+2\binom{n-i-1}{i-1}$ is clearly an integer, and the claim follows.

The integral of each individual monomial will be integral. First we have the identity $$F_n(x)=\sum_{i=0}^{\lfloor(n-1)/2\rfloor}{n-i-1\choose i}x^{n-2i-1},$$ so my claim is that $$\binom{n-i-1}{i}\int_0^1 (k+nz)^{n-2i-1}dz=\binom{n-i-1}{i}\cdot\frac{(k+n)^{n-2i}-k^{n-2i}}{n(n-2i)} \in \mathbb Z.$$ By the binomial theorem we can write $(k+n)^{n-2i}=k^{n-2i}+nk^{n-2i-1}(n-2i)+n^2d$ for some integer $d$. So we can write $$\binom{n-i-1}{i}\cdot\frac{(k+n)^{n-2i}-k^{n-2i}}{n(n-2i)}=k^{n-2i-1}\binom{n-i-1}{i}+d\cdot\frac{n}{n-2i}\binom{n-i-1}{i}$$ it suffices to show that $\frac{n}{n-2i}\binom{n-i-1}{i}$ is an integer. However we can check that $\frac{n}{n-2i}\binom{n-i-1}{i}=\binom{n-i-1}{i}+2\binom{n-i-1}{i-1}$ and the claim follows.

The integral of each individual monomial will be integral. First we have the identity $$F_n(x)=\sum_{i=0}^{\lfloor(n-1)/2\rfloor}{n-i-1\choose i}x^{n-2i-1},$$ so my claim is that $$\binom{n-i-1}{i}\int_0^1 (k+nz)^{n-2i-1}dz \in \mathbb Z.$$ Since $\int_0^1 (k+nz)^{n-2i-1}dz=\frac{(k+n)^{n-2i}-k^{n-2i}}{n-2i}$ and $n$ divides $(k+n)^{n-2i}-k^{n-2i}$, it suffices to show that $$\frac{n}{n-2i}\binom{n-i-1}{i}\in \mathbb Z.$$ However $\frac{n}{n-2i}\binom{n-i-1}{i}=\binom{n-i-1}{i}+2\binom{n-i-1}{i-1}$ is clearly an integer, and the claim follows.

The integral of each individual monomial will be integral. First we have the identity $$F_n(x)=\sum_{i=0}^{\lfloor(n-1)/2\rfloor}{n-i-1\choose i}x^{n-2i-1},$$ so my claim is that $$\binom{n-i-1}{i}\int_0^1 (k+nz)^{n-2i-1}dz \in \mathbb Z.$$ Since $\int_0^1 (k+nz)^{n-2i-1}dz=\frac{(k+n)^{n-2i}-k^{n-2i}}{n(n-2i)}$ and $(k+n)^{n-2i}-k^{n-2i}=nk^{n-2i-1}(n-2i)+n^2d$ for some integer $d$, we have $$\binom{n-i-1}{i}\cdot\frac{(k+n)^{n-2i}-k^{n-2i}}{n(n-2i)}=k^{n-2i-1}\binom{n-i-1}{i}+d\cdot\frac{n}{n-2i}\binom{n-i-1}{i}$$ it suffices to show that $$\frac{n}{n-2i}\binom{n-i-1}{i}\in \mathbb Z.$$ However $\frac{n}{n-2i}\binom{n-i-1}{i}=\binom{n-i-1}{i}+2\binom{n-i-1}{i-1}$ is clearly an integer, and the claim follows.