Types of Numbers

A prime number is defined as a number with exactly two distinct factors: 1 and itself. The number 1 has only one factor — just 1. Since it doesn’t have two distinct factors, it doesn’t meet the definition.

If we allowed 1 to be prime, the Fundamental Theorem of Arithmetic would break — we could write 6 = 2 × 3 but also 6 = 1 × 2 × 3 = 1 × 1 × 2 × 3, and so on. Every number would have infinitely many prime factorisations. Excluding 1 keeps things unique.

An even number is any integer that can be written as \( 2 \times n \) for some integer \( n \). Since \( 0 = 2 \times 0 \), zero fits the definition perfectly. It’s also divisible by 2 with no remainder: \( 0 \div 2 = 0 \).

Looking at the pattern of even and odd numbers: … −4, −2, 0, 2, 4 … — zero sits exactly where the next even number should be. It also ends in 0, which is an even digit. Every test for “evenness” confirms that 0 is even.

Any odd number can be written as “one more than an even number.” So if we add two odd numbers, we’re adding two even numbers plus two extra ones. The two even parts give an even total, and the two extra ones make 2 — which is also even. Even + even = even.

Algebraically, if the two odd numbers are \( 2a + 1 \) and \( 2b + 1 \), their sum is \( 2a + 2b + 2 = 2(a + b + 1) \), which is a multiple of 2 and therefore even. Try it: 3 + 5 = 8, 7 + 9 = 16, 11 + 13 = 24 — always even.

Factors usually come in pairs: for 12, the pair 3 × 4 gives two factors (3 and 4), 2 × 6 gives two more, and 1 × 12 gives two more — six factors in total (even). But for square numbers, one “pair” contains the same number twice. For example, 36 has the pair 6 × 6, but we only count the 6 once.

Check it: 1 has 1 factor, 4 has 3 factors (1, 2, 4), 9 has 3 factors (1, 3, 9), 16 has 5 factors (1, 2, 4, 8, 16), 25 has 3 factors (1, 5, 25). Every time, the square root “unpairs” itself, leaving an odd total. Non-square numbers always have an even number of factors.

Example: 4 — factors are 1, 2, 4

Another: 9 — factors are 1, 3, 9

Creative: 49 — factors are 1, 7, 49. Or 121 (= 11²). The key insight is that numbers with exactly three factors are always the square of a prime number.

Trap: 6 — a student might list the factors as “1, 2, 3” and stop, thinking that’s three. But 6 itself is also a factor, so 6 actually has four factors: 1, 2, 3, and 6.

Example: 9 (= 3 × 3)

Another: 15 (= 3 × 5)

Creative: 77 — many people assume 77 is prime, but 77 = 7 × 11. Or 1 — it’s odd and not prime (since 1 is neither prime nor composite).

Trap: 47 — a student might offer this thinking “it doesn’t feel prime” or confusing it with a composite number, but 47 is prime. It has exactly two factors: 1 and 47.

Example: 13 → 31 (both prime)

Another: 37 → 73 (both prime)

Creative: 71 → 17 (both prime). Or 11 → 11, a palindromic prime that is its own reverse. Note: primes that reverse to a different prime are called emirps (prime spelled backwards) — palindromic primes like 11 don’t count as emirps.

Trap: 19 → 91. While 19 is prime, 91 = 7 × 13 and is not prime. Many students assume 91 is prime because it “looks” prime, but checking divisibility by 7 reveals 91 ÷ 7 = 13.

Example: 64 — it’s \( 8^2 \) and \( 4^3 \)

Another: 1 — it’s \( 1^2 \) and \( 1^3 \)

Creative: 729 — it’s \( 27^2 \) and \( 9^3 \). Or 0, which is \( 0^2 \) and \( 0^3 \). The pattern is that these numbers are always sixth powers: \( n^6 = (n^3)^2 = (n^2)^3 \).

Trap: 16 — a student might think “16 is \( 4^2 \) and \( 2^4 \), so it’s both a square and a cube.” But \( 2^4 \) is a fourth power, not a cube. To be a cube you need three equal factors, and \( 16 \neq n^3 \) for any whole number \( n \).

Some square numbers are even: \( 2^2 = 4 \), \( 4^2 = 16 \), \( 6^2 = 36 \). But others are odd: \( 1^2 = 1 \), \( 3^2 = 9 \), \( 5^2 = 25 \).

The rule is simple: squaring an even number gives an even square, and squaring an odd number gives an odd square. The square inherits the “oddness” or “evenness” of its root.

Almost all primes are odd — 3, 5, 7, 11, 13, 17, 19 … But there is exactly one exception: 2. It has exactly two factors (1 and 2), which makes it prime, and it’s even.

In fact, 2 is the only even prime. Every other even number is divisible by 2, so it has at least three factors (1, 2, and itself) and can’t be prime. This makes 2 unique among primes.

If \( p \) and \( q \) are both prime, then \( p \times q \) has at least three factors: 1, \( p \), and \( q \) (and the product itself makes four unless \( p = q \)). Since it has more than two factors, it cannot be prime.

For example: 2 × 3 = 6 (not prime), 3 × 5 = 15 (not prime), 2 × 2 = 4 (not prime). The product always has the two primes as factors, which immediately disqualifies it from being prime.

An odd number can be written as \( 2a + 1 \) and another as \( 2b + 1 \). Their product is \( (2a+1)(2b+1) = 4ab + 2a + 2b + 1 = 2(2ab + a + b) + 1 \), which is always odd. Try it: 3 × 5 = 15, 7 × 9 = 63, 11 × 13 = 143 — all odd.

This makes a nice contrast with the sum of two odd numbers, which is always even. Students who’ve just learned that “odd + odd = even” may be tempted to think “odd × odd = even” by analogy — but multiplication and addition behave very differently here.

The student is confusing “squaring” with “doubling.” They think “square” means “multiply by 2,” so they’ve produced powers of 2 (2, 4, 8, 16, 32) instead of square numbers.

Squaring means multiplying a number by itself: \( n \times n \). The first five square numbers are \( 1^2 = 1 \), \( 2^2 = 4 \), \( 3^2 = 9 \), \( 4^2 = 16 \), \( 5^2 = 25 \). Notice only 4 and 16 appear in both lists — that’s a coincidence, not a rule.

The answer happens to be correct — 91 is indeed not prime — but the reasoning is completely wrong. The student has invented a rule: “if the digit sum is even, the number can’t be prime.” This rule is false.

A counterexample destroys it: 29 has a digit sum of 11 (odd), and it is prime — consistent with the rule. But 2 has a digit sum of 2 (even), and it is prime — breaking the rule entirely. The real reason 91 isn’t prime is that 91 = 7 × 13. The correct way to check primality is to test for factors, not to examine digit sums.

The student has over-generalised: “all primes are odd.” While it’s true that every prime after 2 is odd, 2 itself is prime. It has exactly two factors — 1 and 2 — which is the definition of a prime number.

2 is the only even prime, and this is precisely what makes it special. The reason all other even numbers fail is that they’re divisible by 2 (giving them at least three factors), but 2 itself can’t be “divided by a smaller even number.” Remembering 2 as the exception is crucial.

The student is confusing square numbers with cube numbers. They correctly found that \( 12 \times 12 = 144 \), which proves 144 is a square number (\( 12^2 \)). But a cube number requires three equal factors: \( n \times n \times n \).

Checking: \( 5^3 = 125 \) and \( 6^3 = 216 \). Since 144 falls between these and isn’t equal to either, 144 is not a cube number. The student needs to distinguish between \( n^2 \) (square) and \( n^3 \) (cube) — the exponent matters!