Recipe

Solve a linear equation over $\mathbb F_p$

Use when: you have an equation like $ax+b=c$ with every number drawn from a finite field $\mathbb F_p$ , not from $\mathbb R$

Goal: the unique solution $x\in\{0,\dots,p-1\}$ satisfying the equation mod $p$

Taught in Vector spaces over ℚ, ℝ, ℂ, and 𝔽ₚ

Quick recipe

Isolate the variable term and reduce
Find the coefficient's multiplicative inverse mod $p$
Multiply both sides by that inverse
Reduce the result into $\{0,\dots,p-1\}$
Check the solution in the original equation, entirely mod $p$

1
Move the constant term to the other side exactly as over $\mathbb R$ , then reduce the result mod $p$ so it lands in $\{0,\dots,p-1\}$ .

Example

For $4x+2=3$ in $\mathbb F_{11}$ : subtracting $2$ from both sides gives $4x=1$ , already in range.

Why: addition and subtraction work exactly the same way in any field, so this step needs no new arithmetic, only a later reduction mod $p$ .
2
Since $p$ is small, test every candidate $1,2,\dots,p-1$ until one multiplies the coefficient to $1$ mod $p$ .

Example

Testing candidates for $4^{-1}$ in $\mathbb F_{11}$ : $4(1)=4$ , $4(2)=8$ , $4(3)=12\equiv1$ . So $4^{-1}=3$ .
- the coefficient is already $\equiv 1 \pmod p$ — it is its own inverse, no search needed.
- the coefficient is any other nonzero residue — search every candidate from $1$ to $p-1$ .
⚡ Shortcut

When the coefficient is $\equiv -1\pmod p$ (that is, $p-1$ ), it is always its own inverse too, since $(-1)(-1)=1$ in any field.

Why: a prime $p$ guarantees every nonzero element has exactly one such partner, the fact that makes $\mathbb F_p$ a field at all.
3
Multiply the isolated constant by the coefficient's inverse from Step 2, staying mod $p$ throughout.

Example

$x = 4^{-1}\cdot 1 = 3\cdot 1 = 3 \pmod{11}$ .

Why: multiplying by the inverse is the field's own version of dividing, the only way to undo multiplication by the coefficient.
4
If the product lands outside that range, keep subtracting or adding $p$ until it doesn't.

Example

$3$ already lies in $\{0,\dots,10\}$ , so no further reduction is needed.

Why: $\mathbb F_p$ 's elements are only these $p$ representatives, so any other integer names the same element as one of them.
5
Substitute the answer back into the original equation and confirm both sides match after reducing mod $p$ .

Example

$4(3)+2=14\equiv 3\pmod{11}$ , matching the right-hand side.

Why: this catches an arithmetic slip in an earlier step before it gets reported as the final answer.

Common mistakes

Forgetting the final reduction mod $p$ and reporting a number outside $\{0,\dots,p-1\}$ .
Confusing the additive inverse (negation) with the multiplicative inverse this recipe needs.
Searching for an inverse of $0$ , which never exists since $0$ has no multiplicative inverse in any field.
Using ordinary real-number division instead of multiplying by the mod- $p$ inverse.

Solve a linear equation over \mathbb F_p

Solve a linear equation over $\mathbb F_p$