Problem 11111111 worked answer

The Same Equation, Different Worlds

Designed for Grade 11 · Grade 12 · Uses HS PRE math · About 20–45 minutes

Complete to collectTanzanite CrystalLevel 2

At a glance

Part 1: Over both the integers and complex numbers, the complete solution set is $𝑋 = 1, - 1$ .
Part 2: Modulo 15, $42 \equiv 1$ , while $(4 - 1) (4 + 1) = 3 \cdot 5 \equiv 0$ although neither factor is 0. The zero-product inference fails.
Part 3: Every matrix $𝐴 𝑡 = (1 𝑡 0 - 1)$ satisfies $𝐴 2 𝑡 = 𝐼$ , is neither $𝐼$ nor $- 𝐼$ , and gives nonzero factors $𝐴 𝑡 - 𝐼$ and $𝐴 𝑡 + 𝐼$ whose product is $𝑂$ . There are infinitely many such matrices.

Key idea: factorization is not the whole proof

The identity

𝑋 2 - 𝟏 = (𝑋 - 𝟏) (𝑋 + 𝟏)

survives in all three settings. The step from a zero product to a zero factor depends on the surrounding mathematical structure.

1. Ordinary numbers

From

𝑋 2 = 1

we obtain

𝑋 2 - 1 = 0

and then factor:

(𝑋 - 1) (𝑋 + 1) = 0 .

The zero-product rule holds for both integers and complex numbers. Therefore

𝑋 - 1 = 0 o r 𝑋 + 1 = 0,

𝑋 = 1 o r 𝑋 = - 1 .

Both candidates work:

12 = 1, (- 1) 2 = 1 .

The factorization and zero-product rule exclude every other integer or complex number. Thus the complete solution set in either world is ${1, - 1}$ .

2. The 15-position world

First check the equation:

42 = 16 \equiv 1 (m o d 15) .

The two factors are

4 - 1 \equiv 3 (m o d 15)

and

4 + 1 \equiv 5 (m o d 15) .

Neither factor is congruent to 0 modulo 15, but

3 \cdot 5 = 15 \equiv 0 (m o d 15) .

Therefore the factorization

(4 - 1) (4 + 1) \equiv 0 (m o d 15)

is valid. What fails is the next inference: a product congruent to 0 modulo 15 does not force either factor to be congruent to 0 modulo 15.

This explains how $4$ can solve the equation without being congruent to either $1$ or $- 1$ . The supplied complete list $1, 4, 11, 14$ is consistent with this failure.

3. The matrix world

Square $𝐴 𝑡$ :

𝐴 2 𝑡 = (1 𝑡 0 - 1) (1 𝑡 0 - 1) = (1 𝑡 - 𝑡 01) = (1001) = 𝐼 .

Thus every real $𝑡$ gives a matrix solution.

The diagonal entries of $𝐴 𝑡$ are $1$ and $- 1$ . They are not the two diagonal entries of $𝐼$ , and they are not the two diagonal entries of $- 𝐼$ . Therefore

𝐴 𝑡 \neq 𝐼, 𝐴 𝑡 \neq - 𝐼

for every real $𝑡$ .

If $𝑡 1 \neq 𝑡 2$ , the upper-right entries of $𝐴 𝑡 1$ and $𝐴 𝑡 2$ differ, so the matrices are distinct. Since there are infinitely many real values of $𝑡$ , the equation $𝑋 2 = 𝐼$ has infinitely many matrix solutions.

Now compute the two factors:

𝐴 𝑡 - 𝐼 = (0 𝑡 0 - 2), 𝐴 𝑡 + 𝐼 = (2 𝑡 00) .

The first factor is nonzero because its lower-right entry is $- 2$ . The second is nonzero because its upper-left entry is $2$ . Yet

(0 𝑡 0 - 2) (2 𝑡 00) = (0000) = 𝑂 .

The difference-of-squares factorization itself is also valid:

(𝐴 𝑡 - 𝐼) (𝐴 𝑡 + 𝐼) = 𝐴 2 𝑡 + 𝐴 𝑡 𝐼 - 𝐼 𝐴 𝑡 - 𝐼 2 = 𝐴 2 𝑡 - 𝐼,

because $𝐴 𝑡 𝐼 = 𝐼 𝐴 𝑡 = 𝐴 𝑡$ and $𝐼 2 = 𝐼$ .

Check: what crossed the folds

The factorization into $𝑋 - 𝟏$ and $𝑋 + 𝟏$ stayed valid in every fold. The zero-product rule held for integers and complex numbers but failed modulo 15 and for matrices.

Nonzero objects whose product is zero are called zero divisors. The 15-position world and the matrix world contain zero divisors; the integer and complex-number worlds do not.

Technical fit and rating

MJ HS:PRE.5 · C2 · W3 Standard challenge · Heavy workload

Algebraic structure and equations · Number and quantity

How the rating works →