Question:
Find $x$ such that the following properties are true or prove that no such $x$ exists.
Let:
- $x>0$ be an integer
- $p_1, p_2, p_3$ be distinct odd primes
- $1 \le a < p_3$ is an integer
Find $x$ with the following properties:
- $x \equiv p_2 \pmod {p_3}$
- $x \equiv p_3 \pmod {p_1}$
- $x = p_1 + ap_2$
Here are my observations:
For any specific $p_1, p_2, p_3$, the problem is straight forward. Use the Chinese Remainder Theorem to solve for:
- $x \equiv p_1 \pmod {p_2}$
- $x \equiv p_2 \pmod {p_3}$
- $x \equiv p_3 \pmod {p_1}$
Then see if the minimal $x$ has the form $p_1 + ap_2$. In all the cases I am testing, $x > p_1 + (p_3 - 1)p_2$ in which case no such $a$ exists.
For example:
- $x \equiv 3 \pmod 5$
- $x \equiv 5 \pmod 7$
- $x \equiv 7 \pmod 3$
The minimal $x$ is $103 > 3 + 5\times 7 = 38$
So, in this case, no such $a$ exists.
I am having troubling proving that no solution exists and I am not able to find example where it is true.
You don't mention any specific ordering requirements among the primes, so consider $p_1 = 7$, $p_2 = 5$ and $p_3 = 3$. You then get
$$x \equiv p_1 \pmod{p_2} \implies x \equiv 7 \pmod{5} \tag{1}\label{eq1A}$$
$$x \equiv p_2 \pmod{p_3} \implies x \equiv 5 \pmod{3} \tag{2}\label{eq2A}$$
$$x \equiv p_3 \pmod{p_1} \implies x \equiv 3 \pmod{7} \tag{3}\label{eq3A}$$
You can easily confirm that $x = 17$ satisfies the $3$ equations above, and it's the smallest positive integer that does. Also,
$$p_1 + (p_3 - 1)p_2 = 7 + (3 - 1)5 = 17 = x \tag{4}\label{eq4A}$$
so $a = p_3 - 1 = 2$ in this case satisfies your requirements.
I haven't checked, but I'm fairly certain counter-examples will exist even if you impose an ordering requirement on the primes, e.g., $x_1 \lt x_2 \lt x_3$.