Four Ways Back to One

At a glance

• Part 1: Modulo 3, the square remainders are 0,1,1, so the inputs are 1,2. Modulo 5, the square remainders are 0,1,4,4,1, so the inputs are 1,4.
• Part 2: The four dial positions are 1,4,11,14.
• Part 3: Their squares are 15(0) +1, 15(1) +1, 15(8) +1, and 15(13) +1. The two choices modulo 3 and two choices modulo 5 exhaust every possible solution.

Key idea: split, then recombine

If a square leaves remainder 1 after division by 15, it must also leave remainder 1 after division by each factor 3 and 5. The two small dials reveal two possible remainders apiece. Recombining those choices produces four cases, not fifteen unrelated trials.

1. Build the two clues

Modulo 3:

The completed row is 0,1,1, and the circled inputs are 1,2.

Modulo 5:

The completed row is 0,1,4,4,1, and the circled inputs are 1,4.

2. Recombine the clues

The positions with remainder 1 modulo 3 are 1,4,7,10,13. The positions with remainder 2 modulo 3 are 2,5,8,11,14.

The positions with remainder 1 modulo 5 are 1,6,11. The positions with remainder 4 modulo 5 are 4,9,14.

Intersecting the appropriate lists completes the table:

Therefore the complete solution list is

3. Check and prove completeness

Each listed position works:

For completeness, suppose 𝑥2 ≡1 (mod 15). Then 15 divides 𝑥2 −1, so both 3 and 5 divide 𝑥2 −1. The small tables prove that 𝑥 must have remainder 1 or 2 modulo 3 and remainder 1 or 4 modulo 5. Those are exactly the four rows of the recombination table.

Each row contains exactly one representative from 0 through 14, and the direct checks show that all four representatives work. Therefore no fifth dial position can satisfy the equation.

Teaching note

The four rows are a small, concrete instance of the Chinese remainder principle: a position modulo 15 is determined by its paired remainders modulo 3 and modulo 5. The worksheet does not require that theorem by name.