similar to http://www.mathlinks.ro/viewtopic.php?t=1366 Let $ n!=p*q$, where $ p$ shares prime factors with $ a$ and $ (q,a)=1$. Then we can find $ t>1$ such that $ q|a^t-1=b$. Also $ p|a^p$. $ P(x)=2*a^p(1+\sum_{i=1}^n\frac{x(x-1)\cdots(x-i+1)}{i!}b^i)+3$. Since $ i!|n!=pq|a^p*b^i$, the coefficients are all integers. For any $ 0\le m\le n$, $ P(m)=2*a^p(1+\sum_{i=1}^mC_m^ib^i)+3=2*a^p*(b+1)^m+3=2*a^{p+tm}+3$.

xxp2000 wrote: similar to http://www.mathlinks.ro/viewtopic.php?t=1366 Let $ n! = p*q$, where $ p$ shares prime factors with $ a$ and $ (q,a) = 1$. Then we can find $ t > 1$ such that $ q|a^t - 1 = b$. Also $ p|a^p$. $ P(x) = 2*a^p(1 + \sum_{i = 1}^n\frac {x(x - 1)\cdots(x - i + 1)}{i!}b^i) + 3$. Since $ i!|n! = pq|a^p*b^i$, the coefficients are all integers. For any $ 0\le m\le n$, $ P(m) = 2*a^p(1 + \sum_{i = 1}^mC_m^ib^i) + 3 = 2*a^p*(b + 1)^m + 3 = 2*a^{p + tm} + 3$. xxp2000 wrote: $ q|a^t - 1 = b$ What does this mean? Do you mean $ qb=a^t-1$? If so, then how does: xxp2000 wrote: $ 2*a^p*(b + 1)^m + 3 = 2*a^{p + tm} + 3$ ?? i.e. how does $ (b + 1)^m=a^{tm}$? Because obviously $ b\not= a^t-1$.

$ q|a^t-1=b$ means $ a^t-1=b$ and $ b$ is multiple of $ q$.

It's simple using Language's interpolation formula.Assume $$P\left( i\right) =2^{k_{i}}+3$$then$$P\left( x\right) =\sum ^{n}_{i=0}P\left( i\right) \cdot\prod _{j\neq i}\dfrac {\left( x-j\right) }{\left( i-j\right) }$$For a big $k$,$$f\left( x\right) =\sum ^{n}_{i=0}\left( 2a^{k}+3\right) \cdot \prod _{j\neq i}\dfrac {\left( x-j\right) }{\left( i-j\right) }=2a^{k}+3$$So we just need to make $$\sum ^{n}_{i=0}2\left( a^{k_{i}}-a^{k}\right) \prod _{j\neq i}\dfrac {\left( x-j\right) }{\left( i-j\right) }$$a integral polynomial,thus we just need to make $$n! \vline a^{k_{i}}-a^{k}$$,it's easy just make $k$ big enough and $\varphi \left( n!\right) \vline k_{i}-k$

P-H-David-Clarence wrote: It's simple using Language's interpolation formula.Assume $$P\left( i\right) =2^{k_{i}}+3$$then$$P\left( x\right) =\sum ^{n}_{i=0}P\left( i\right) \cdot\prod _{j\neq i}\dfrac {\left( x-j\right) }{\left( i-j\right) }$$For a big $k$,$$f\left( x\right) =\sum ^{n}_{i=0}\left( 2a^{k}+3\right) \cdot \prod _{j\neq i}\dfrac {\left( x-j\right) }{\left( i-j\right) }=2a^{k}+3$$So we just need to make $$\sum ^{n}_{i=0}2\left( a^{k_{i}}-a^{k}\right) \prod _{j\neq i}\dfrac {\left( x-j\right) }{\left( i-j\right) }$$a integral polynomial,thus we just need to make $$n! \vline a^{k_{i}}-a^{k}$$,it's easy just make $k$ big enough and $\varphi \left( n!\right) \vline k_{i}-k$ "$For a big $k$,$$f\left( x\right) =\sum ^{n}_{i=0}\left( 2a^{k}+3\right) \cdot \prod _{j\neq i}\dfrac {\left( x-j\right) }{\left( i-j\right) }=2a^{k}+3$ So we just need to make $\sum ^{n}_{i=0}2\left( a^{k_{i}}-a^{k}\right) \prod _{j\neq i}\dfrac {\left( x-j\right) }{\left( i-j\right) }$a integral polynomial" What do these 2 lines mean? Can you explain, please. Also the last line $\varphi \left( n!\right) \vline k_{i}-k$. I think you use Euler theorem but if you use like that, we need $(a,n!)=1$.