All of the prime numbers s.t. $3x^p+px^3\equiv \,\,0\,\, (\text{mod} \,\, 5)$ has solutions $\forall x \in \mathbb{Z}$

Question

Determine all of the prime numbers $\,\,p\,\,\,$ s.t. $\,\,\,2 \le p \le 50\,\,\,$s.t. the following congruence has solutions $\forall x \in \mathbb{Z}$

$$3x^p+px^3\equiv \,\,0\,\, (\text{mod} \,\, 5)$$

I am bogged down with this exercise, any hint? Maybe the little Fermat's theorem (?) on $x(3x^{p-1}+px^2)\equiv \,\,0\,\, (\text{mod} \,\, 5)$.

Answer 1

As lulu says, the equation

$$3x^p + px^3 \equiv 0 \pmod 5 \tag{1}\label{eq1}$$

is always true when $x \equiv 0 \pmod 5$, so it doesn't do anything to limit which possible values to check. Next, try $x \equiv 1 \pmod 5$. This gives

$$3 + p \equiv 0 \pmod 5 \Rightarrow p \equiv 2 \pmod 5 \tag{2}\label{eq2}$$

Within your provided range, this means that the only possible choices are

$$p = 2, 7, 17, 37, 47 \tag{3}\label{eq3}$$

Next, consider $x \equiv 4 \equiv -1 \pmod 5$. For $p = 2$, this gives

$$3 - 2 \equiv 1 \pmod 5 \tag{4}\label{eq4}$$

However, for any odd $p$, \eqref{eq1} becomes

$$-3 - p \equiv -5 \equiv 0 \pmod 5 \tag{5}\label{eq5}$$

Thus, it'll always work then. This leaves just checking $2$ and $3$. For $x \equiv 2 \pmod 5$, we get

$$3 \times 2^p + \left(5 + 3\right)p \equiv 0 \Rightarrow 2^p + p \equiv 0 \pmod 5 \tag{6}\label{eq6}$$

As $p \equiv 2 \pmod 5$, \eqref{eq6} becomes

$$2^p \equiv 3 \pmod 5 \tag{7}\label{eq7}$$

This is true iff $p \equiv 3 \pmod 4$. Among the remaining values in \eqref{eq3} for $p$, this leaves just $p = 7, 47$. Finally, for checking $x \equiv 3$, note that $3 \equiv -2 \pmod 5$. As the powers of $x$ in \eqref{eq1} are odd (i.e., $p$ and $3$), this means the resulting value would be the negative of that obtained for $x \equiv 2 \pmod 5$, so the same values of $p$ will work with it as well.

In summary, the only values of $p$ which work are $7$ and $47$.

Answer 2

Assume $p>2$ (you can check explicitly that $p=2$ is not a solution). Let's work with $p \pmod 4$

Case I: $p\equiv 1 \pmod 4$.

Then $x^{p}\equiv x\pmod 5$ so your congruence reads $$3x\equiv -px^3\pmod 5$$ but it is clear that this can not hold for all $x$.

Case II: $p\equiv 3\pmod 4$

Then $x^p\equiv x^3\pmod 5$ so your congruence reads $$3x^3\equiv -px^3\pmod 5$$ and that will always hold if we have $p\equiv 2 \pmod 5$

It follows that we want to solve the simultaneous congruences $$p\equiv 3 \pmod 4 \quad \&\quad p\equiv 2 \pmod 5$$

Straightforward calculations show that this has the solution $$p\equiv 7 \pmod {20}$$ so in your range, the odd solutions are $$\boxed {7,47}$$

Answer 3

Obviously it has to be valid for every whole number, so lets start with $x=1$ and $ x=-1$.
For $x=1$ we get $3+p \equiv 0 $ ($mod$ $5$), therefore $p \in $ {$2,7,17,37,47$}, and from $x=-1$ it follows that $p=2$ is not possible.
So we can wrtie $p$ in the form $p=7+10k$, where $k \in$ {$0,1,3,4$}. Now our equation has the form $ x^3(3x^{4+10k}+7+10k)\equiv0 $ ($mod$ $5$) which is:
$ x^3(3x^{4+10k}+2)\equiv0 $ ($mod$ $5$).
It's clear that $x^3$ is not always divisible by 5, so we can distinguish two cases:

1) $x=5m$
2) $x\neq5m$

There is nothing to prove in the first one, so let's focus on the second case. We have $(3x^{4+10k}+2)\equiv0 $ ($mod$ $5$) so it follows $x^{4+10k}\equiv1 $ ($mod$ $5$). Fermat's little theorem tells us this is always valid for $k=0$ and $k=4$ and for $k=1$ and $k=3$ you should be able to find some counter examples. So at the end we get two solutions - $p=7$ and $p=47$