Problem $43$ on PDF-Page $20$ of: https://blngcc.files.wordpress.com/2008/11/rmc2007.pdf
Let $G$ be a finite group and $p_1, p_2, \ldots, p_n$ be distinct prime divisors of $|G|$, such that for each $p_i$ there exists $x_i, y_i$ in $G$ with $$\operatorname{ord}\left(x_i\right)=\operatorname{ord}\left(y_i\right)=p_i$$ and $y_i$ is not a power of $x_i$.
Prove that
$$
|G| \geq 1+\prod_{i=1}^n\left(p_i^2-1\right).
$$
Update:
$S_4$ is a counterexample as
$$
|S_4|=4!=24<25=1+(2^2-1)(3^2-1)
$$
and all the conditions for $S_4$ are fulfilled.
I suggest a new version of the Problem:
Let $G$ be a finite group and $p_1, p_2, \ldots, p_n$ be distinct prime divisors of $|G|$, such that for each $p_i$ there exists $x_i, y_i$ in $G$ with $$\operatorname{ord}\left(x_i\right)=\operatorname{ord}\left(y_i\right)=p_i$$ and $y_i$ is not a power of $x_i$.
Prove that
$$
|G| \geq 1+\sum_{i=1}^n\left(p_i^2-1\right).
$$
Best Answer
Here is a sketch proof of the amended version of the problem with the product replaced by sum. That is $$|G| \ge 1 + \sum_{i=1}^n(p_i^2-1).$$ It is enough to prove that the for each $i$ the number of $p_i$-elements of $G$ is at least $p_i^2-1$. This is clear if a Sylow $p_i$-subgroup $P_i$ of $G$ has order at least $p_i^2$, so we just have to consider the case when $|P_i|=p_i$.
By the assumption that we have elements $x_i$ and $y_i$ of order $p_i$, where $x_i$ is not a power of $y_i$, $P_i$ cannot be the unique subgroup of $G$ of order $p$, so by Sylow's Theorem $G$ has at least $p_i+1$ Sylow $p_i$-subgroups, which together contain at least $(p_i+1)(p_i-1)=p_i^2-1$ elements of order $p_i$.
I think the only examples in which we get equality are $A_4$ and groups $C_p \times C_p$ for primes $p$, but I have not proved that.