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suppose the polynomial with integer coefficients $p(x) = x^4+a_3x^3+a_2x^2+a_1x+a_0$ has four real roots and is such that $24 | p(n)$ for all natural numbers $n$. Prove there are infinitely many natural numbers such that $p(n)$ is not divisible by 5.

I have figured that having one number $k \in \mathbb{N}$ such that 5 does not divide $p(k)$ implies there are infinitely many as $p(k) \equiv p(k+5s)$ (mod 5), $s \in \mathbb{N}$. It is also equivalent to proving there can't exist a 4th degree monic polynomial which is always divisible by 120 for natural inputs. Furthermore, I don't know how the real roots condition has anything to do with the result. Any help would be appreciated.

Teddy38
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Lorenzo Catani
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  • If $a_i\in\mathbb{Z}$ (which is only stated in the title but not in the question), then just reduce $P$ mod 5 and consider $\bar{P}\in\mathbb{F}_5[x]$. What do you know about roots of $\bar{P}$? – user10354138 Sep 08 '21 at 12:03
  • thank you, I have edited the question so as to make the integer condition more explicit. I think I understand now: $\bar{P}$ has 5 roots (all the residues mod 5) but is only of degree 4, therefore must be the zero polynomial (which is in contradiction with the hypothesis). However, does this mean the $24 | p(n)$ condition is irrelevant to the result, or am I missing something? – Lorenzo Catani Sep 08 '21 at 12:51
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    Yes, both $24\mid p(n)$ and $p$ having only real roots are irrelevant in this proof. Perhaps they were seeking another proof where this would come up (e.g., we can calculate the fourth forward difference $(\Delta^4 p)(n)=p(n+4)-4p(n+3)+6p(n+2)-4p(n+1)+p(n)=24$ and that is incompatible with $120\mid p(n)$ for all $n$, but this proof still wouldn't need the only real roots assumption). – user10354138 Sep 08 '21 at 13:13
  • This result implies that $\gcd(p(n)) \mid 4!=24$, so we cannot have $5\mid \gcd(p(n))$. Having real roots or being divisible by $24$ is irrelevant to the fact $5 \not\mid \gcd(p(n))$, it will hold for any monic 4th degree polynomial in $\mathbb{Z}[x]$. – Sil Nov 21 '21 at 22:01

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Let's suggest that $120|p(n)$ for all natural n. Taking $n=120b-2,120b-1,120b,120b+1,120b+2$, where $b$ is natural number, one can get that 120 should divide $16-8a_3+4a_2-2a_1+a_0$, $1-a_3+a_2-a_1+a_0$, $a_0$, $1+a_3+a_2+a_1+a_0$, $16+8a_3+4a_2+2a_1+a_0$. Combining by adding and subtracting gives $120|a_0,2+2a_0+2a_2,32+2a_0+8a_2$. But these three numbers $c_1=a_0$, $c_2=2+2a_0+2a_2$, $c_3=32+2a_0+8a_2$ cannot be divisible by 120 simultaneously because $c_3-4c_2+6c_1=24$ which is not divisible by 120.

Ivan Kaznacheyeu
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