No Formula Needed: Albert Finds Bézout His Way

Yesterday, Albert and I took a deep dive into a simple-looking but powerful topic: solving equations of the form

mx + ny = k
where m, n, k are integers, and x, y are unknown integers.

We started with a warm-up:

3x + 5y = 1

Albert quickly began applying the Euclidean algorithm:

5 = 1×3 + 2
3 = 1×2 + 1
2 = 2×1 + 0

So the gcd is 1. That means there is a solution. He worked backwards and found one:
x = 2, y = -1

From there, he formed the general idea:
An equation of the form mx + ny = k has integer solutions if and only if gcd(m, n) divides k.

Then he challenged himself with a harder one:

100x + 200y = 99

He analyzed it logically:
gcd(100, 200) = 100
99 is not divisible by 100
Therefore, no solution exists

Then came the special moment. He tried to write:

100(x + 2y) = i
so x + 2y = i / 100

I gently pointed out:
Did you realize this 100 is not the same as the 100 from the previous line?

He paused. Thought. Then replaced it with:

x + 2y = i / gcd(100, 200)

That was the kind of moment every math teacher lives for — when a student sees the structure behind the numbers.

He continued thinking aloud, and then generalized:

Let g = gcd(m, n)
If 0 < k < g, then there is no solution.

He even introduced new variables:

Let m = sg, n = rg
Then the equation becomes:
sgx + rgy = k
which means g(sx + ry) = k
so sx + ry = k / g

And he concluded:
Since k/g is not an integer, there's no solution
But sx + ry must be an integer — contradiction

Finally, he wrote:

We can humbly guess:
q = min(|mx + ny|) = gcd(m, n)

He later posted his thoughts in our kids’ Slack channel — raw, real, and logical.

He didn’t memorize anything. He didn’t recite theorems.
He discovered Bézout’s identity step by step.

And I had the privilege of walking right next to him.


Étienne Bézout (1730–1783) was a French mathematician known for Bézout’s identity, which says that the smallest positive value of mx + ny is exactly gcd(m, n).