Prime Numbers

A prime number must have exactly two distinct factors: 1 and itself. The number 1 only has one factor — itself. Since it doesn’t meet the “exactly two factors” requirement, it is not prime.

There’s also a deeper reason: if 1 were considered prime, the Fundamental Theorem of Arithmetic (that every whole number greater than 1 has a unique prime factorisation) would break down — because you could always multiply by extra 1s and get different-looking factorisations.

The factors of 2 are 1 and 2 — exactly two distinct factors. That is the definition of a prime number, so 2 is prime. It doesn’t matter that 2 is even.

In fact, 2 is the only even prime number. Every other even number has 2 as a factor (in addition to 1 and itself), giving it at least three factors. This makes 2 unique — and the reason students often think “primes must be odd” is because 2 is the sole exception.

Suppose you had a “complete” list of primes: 2, 3, 5, 7, 11, … up to some largest prime \(p\). Now multiply them all together and add 1. The result isn’t divisible by any prime on your list (dividing by any of them always leaves remainder 1). So this new number is either prime itself, or has a prime factor that wasn’t on the list.

Either way, the list was incomplete. No matter how long a list of primes you make, there’s always another one missing. This argument was first given by Euclid around 300 BC and is one of the most famous proofs in all of mathematics.

If \(n = a \times b\), then \(a\) and \(b\) can’t both be bigger than \(\sqrt{n}\) — because if they were, \(a \times b > \sqrt{n} \times \sqrt{n} = n\), which is a contradiction. So at least one factor must be \(\leq \sqrt{n}\).

For example, to test whether 97 is prime, \(\sqrt{97} \approx 9.8\), so you only need to check primes up to 9: that’s 2, 3, 5, and 7. None of them divide 97, so it’s prime. Without this shortcut, you’d waste time checking much larger numbers unnecessarily.

A prime number must have exactly two distinct factors. The number 0 can be divided by any non-zero integer without leaving a remainder (\(0 \div 1 = 0\), \(0 \div 2 = 0\), \(0 \div 99 = 0\), etc.).

This means 0 has an infinite number of factors! Since it has infinitely many factors rather than exactly two, it completely breaks the fundamental definition and cannot possibly be a prime number.

Example: 23

Another: 31

Creative: 97 — the largest two-digit prime. Or 101 — pushing into three-digit territory.

Trap: 27 — it’s odd and doesn’t end in 0 or 5, so it looks prime. But 27 = 3 × 9. Students who only check for divisibility by 2 and 5 will miss this.

Example: 9 (= 3 × 3)

Another: 15 (= 3 × 5)

Creative: 77 — it looks like it could be prime, but 77 = 7 × 11. Many students check 2, 3, and 5 then stop, missing the factor of 7.

Trap: 11 — it’s odd, but it is prime (its only factors are 1 and 11), so it doesn’t satisfy the condition. A student who assumes all odd numbers are composite — or who doesn’t bother checking for primality — might offer this.

Example: 5 and 7

Another: 11 and 13

Creative: 41 and 43 — a pair of twin primes that most students won’t think to check. Or 71 and 73.

Trap: 13 and 15 — they are consecutive odd numbers, but 15 = 3 × 5 is not prime. Students who assume all odd numbers are prime will fall into this trap.

Example: 4 (factors: 1, 2, 4)

Another: 9 (factors: 1, 3, 9)

Creative: 169 = 13² (factors: 1, 13, 169). Or 121 = 11². The pattern? Numbers with exactly 3 factors are always the square of a prime.

Trap: 6 — a student might list the factors as 1, 2, 3 and count three, but they’ve forgotten to include 6 itself. The actual factors of 6 are 1, 2, 3, and 6 — that’s four, not three. Forgetting to include the number itself as a factor is a very common error.

Example: 8 and 9. Both are composite (8 has factors 1, 2, 4, 8; 9 has factors 1, 3, 9), but their prime factors (2 and 3) are completely different.

Another: 14 and 15 (Prime factors are 2, 7 and 3, 5).

Creative: 25 and 36 (Prime factors are 5 and 2, 3).

Trap: 15 and 21. Both are composite, but they share the prime factor of 3 (\(15 = 3 \times 5\), \(21 = 3 \times 7\)).

It’s true for many numbers: 7, 11, 13, 17, 19, 23 all have no factor of 2, 3, or 5 — and they’re all prime. Checking these three primes catches a lot of composites.

But it fails for numbers like 49 = 7 × 7 and 77 = 7 × 11. Neither is divisible by 2, 3, or 5, yet neither is prime. This is the “incomplete factor check” misconception — ruling out a few small primes is not enough to confirm primality. You need to test all primes up to the square root.

When both primes are odd, their sum is even: 3 + 5 = 8, 7 + 11 = 18. Odd + odd always gives an even result.

But when one of the primes is 2 (the only even prime), we get odd + even = odd: for example, 2 + 3 = 5, 2 + 7 = 9. So the sum is only guaranteed to be even when neither prime is 2. That sneaky 2 strikes again!

A square number \(n^2\) (where \(n \geq 2\)) always has at least three factors: 1, \(n\), and \(n^2\). That’s already too many for it to be prime. For example, 4 has factors 1, 2, 4; 9 has factors 1, 3, 9; 25 has factors 1, 5, 25.

What about \(1^2 = 1\)? That’s not prime either, because 1 has only one factor. So no square number — not even 1 — is prime.

Every whole number falls into one of six categories: \(6k\), \(6k+1\), \(6k+2\), \(6k+3\), \(6k+4\), or \(6k+5\). Of these, \(6k\) is divisible by 6; \(6k+2\) and \(6k+4\) are even; and \(6k+3\) is divisible by 3. None of these can be prime (for values greater than 3).

That leaves only \(6k+1\) and \(6k+5\) — which are exactly “one more” and “one less” than a multiple of 6. Check it: 5 = 6−1, 7 = 6+1, 11 = 12−1, 13 = 12+1, 17 = 18−1, 19 = 18+1. Note: not every number of the form \(6k \pm 1\) is prime (e.g. 25 = 24+1), but every prime greater than 3 is of this form.

The student stopped checking too early. 91 = 7 × 13, so it is not prime. This is the classic “incomplete factor check” misconception — the student only tested the factors they were comfortable with (2, 3, 4, 5, 6) and skipped 7.

To properly test 91, you need to check all primes up to √91 ≈ 9.5, which means testing 2, 3, 5, and 7. Dividing by 7 gives 91 ÷ 7 = 13 exactly. The lesson: checking only “easy” factors is not enough.

The answer is correct — 47 is prime — but the reasoning is dangerously wrong. The student has invented a rule: “odd + doesn’t end in 0 or 5 = prime.” This is the “ending digit determines primality” misconception.

Counterexamples break the rule immediately: 21 (= 3 × 7), 33 (= 3 × 11), 49 (= 7 × 7), and 77 (= 7 × 11) are all odd, don’t end in 0 or 5, but are not prime. The last digit can rule out factors of 2 and 5 — a useful first step — but it tells you nothing about divisibility by 3, 7, 11, or any other prime. Confirming primality requires systematically checking all primes up to the square root.

The student is confusing “having prime factors” with “being prime”. These are completely different things. A prime number has exactly two factors: 1 and itself. A composite number has more than two factors. Every whole number greater than 1 is one or the other — never both.

6 has four factors (1, 2, 3, 6), so it is composite. The fact that some of its factors happen to be prime (2 and 3) is irrelevant to its own classification. By the student’s logic, almost every number would be “both” — since nearly every composite number has prime factors!

The student found a valid factor pair (5 × 6 = 30), but 6 is not prime — it’s 2 × 3. This is the “confusing any factor with a prime factor” misconception. Prime factorisation requires breaking down every factor until all of them are prime.

The correct prime factorisation is 2 × 3 × 5. A factor tree helps: 30 → 5 × 6, then 6 → 2 × 3. You keep splitting until every branch ends at a prime.