Number of solutions a + b + c ≡ 0 (mod n)

[hxthanh]

Number of solutions

Le $n$ be a positive integer $(n\ge 6)$
How many triad $(a,b,c)$ are integers satisfying condition
$\begin{cases}a+b+c\equiv 0\pmod n\\ 1\le a<b<c\le n \end{cases}\quad$?

[sp]
Result: $\left\lceil\dfrac{(n-1)(n-2)}{6}\right\rceil$

*note: $\lceil x\rceil$ is Ceilling Function (The least integer greater than or equal to $x$)
[/sp]

 

[jvicens]

Hello,

Thank you for your question. The number of solutions to this problem can be determined using a mathematical approach. First, we can rewrite the condition as $a+b\equiv -c \pmod n$. This means that for every fixed value of $c$, there will be exactly one pair of values for $a$ and $b$ that satisfy the condition.

Now, let's consider the possible values of $c$. Since $a<b<c$, we know that $a$ and $b$ can take on values from $1$ to $c-1$. This means that $c$ can take on values from $3$ to $n$. Therefore, the number of possible values for $c$ is $n-2$.

Next, we can determine the number of possible pairs for $a$ and $b$ for a fixed value of $c$. This can be done using the formula for the number of combinations, which is $\binom{n}{2}=\frac{n(n-1)}{2}$. However, we need to subtract $1$ from this value to account for the fact that $a$ and $b$ cannot be equal. This gives us $\frac{n(n-1)}{2}-1$ possible pairs for $a$ and $b$.

Therefore, the total number of solutions is $(n-2)\left(\frac{n(n-1)}{2}-1\right)$. However, this counts each solution twice (once for $a$ and $b$ and once for $b$ and $a$), so we need to divide by $2$ to get the final answer of $\frac{(n-2)(n^2-n-2)}{4}$. Using the ceiling function, we get $\left\lceil\frac{(n-2)(n^2-n-2)}{4}\right\rceil=\left\lceil\frac{n^3-3n^2+2n+4}{4}\right\rceil$.

I hope this helps explain the solution to your problem. Let me know if you have any further questions or if you need me to clarify anything.