find m if $m = p_1^2p_2^2$ and $\varphi(m) = 11424$

Question

$m = p_1^2p_2^2$ and $\varphi(m) = 11424$ where $p_1$ and $p_2$ are prime numbers. I need to find m. I know that ($p_1, p_2$) = 1. Consequently $\varphi(m) = p_1^2p_2^2(1-\dfrac{1}{p_1})(1-\dfrac{1}{p_2}) = 11424$. Also I tried writing 11424 in canonical form: $11424 = 2^5 * 3 * 7 * 17$. What I can notice is that $11424 = 2^5 * 3 * 7 * 17 = 2^5(2^2-1)(2^3-1)(2^4+1)$. But this won't help me either. So I have no idea how to solve this.

Answer 1

Notice that $$\varphi(m) = p_1p_2(p_1-1)(p_2-1)$$ W.l.o.g. we can assume that $p_1>p_2$. Since $\varphi(m) = 2^5 \cdot 3 \cdot 7 \cdot 17$ it follows that $17 \big| p_1p_2(p_1-1)(p_2-1)$, so $17 \le p_1$. But $p_1$ can't be greater than $17$, so $p_1 = 17$. Then, since $7 \big| 17\cdot p_2 \cdot 16\cdot(p_2-1)$ it follows that $7 \le p_2$. But for $p_2 = 7$ we have that $\varphi(m) = 2^5 \cdot 3 \cdot 7 \cdot 17$.
Thus, $m = 7^2 \cdot 17^2$.