Algebraic Proof

Let the three consecutive integers be \( n \), \( n + 1 \), and \( n + 2 \). Their sum is \( n + (n + 1) + (n + 2) = 3n + 3 = 3(n + 1) \). Since \( 3(n + 1) \) is 3 multiplied by an integer, it is always a multiple of 3.

A common mistake is to test specific numbers — for example, \( 1 + 2 + 3 = 6 \) and \( 4 + 5 + 6 = 15 \) — and claim this proves the result. This is the “verification is not proof” error: numerical examples can support a conjecture, but they do not prove it works for every set of three consecutive integers. The algebraic approach above proves it in full generality.

Let the consecutive integers be \( n \) and \( n + 1 \). The difference between their squares is \( (n + 1)^2 \; – \; n^2 = n^2 + 2n + 1 \; – \; n^2 = 2n + 1 \). Since \( 2n + 1 \) is one more than an even number, it is always odd.

Students sometimes struggle with the expansion of \( (n + 1)^2 \), writing it as \( n^2 + 1 \) rather than \( n^2 + 2n + 1 \). This is the “squaring a sum” error — believing \( (a + b)^2 = a^2 + b^2 \). Getting the expansion right is the key step in this proof.

Factorise: \( n^2 + n = n(n + 1) \). Since \( n \) and \( n + 1 \) are consecutive integers, one of them must be even. An even number multiplied by any integer gives an even result, so \( n(n + 1) \) is always even. Students who don’t spot the consecutive-integer structure may try to test individual values instead of using the factorisation, which is the key insight.

An alternative approach is to consider cases. If \( n \) is even, then \( n^2 \) is even and \( n \) is even, so their sum is even. If \( n \) is odd, then \( n^2 \) is odd and \( n \) is odd, so their sum is odd + odd = even. Either way, the result is even.

Expand: \( (2n + 1)^2 = 4n^2 + 4n + 1 \). Factorise the first two terms: \( 4n^2 + 4n + 1 = 2(2n^2 + 2n) + 1 \). Since \( 2n^2 + 2n \) is an integer, the expression \( 2(\text{integer}) + 1 \) is one more than an even number, so it is always odd.

Many students correctly expand to \( 4n^2 + 4n + 1 \) but then stop, claiming “you can see it’s odd because of the +1.” This is the “incomplete proof” error — the proof requires showing why adding 1 to \( 4n^2 + 4n \) always produces an odd number. The key step is rewriting \( 4n^2 + 4n \) as \( 2(2n^2 + 2n) \) to make its evenness explicit.

Example: \( 2n \) — this is 2 times an integer, so always even.

Another: \( 4n + 6 \) — this equals \( 2(2n + 3) \), which has a factor of 2, so always even.

Creative: \( n(n \; – \; 1) \) — this is the product of two consecutive integers (one must be even), so it is always even.

Trap: \( n + 2 \) — a student might think “adding 2 makes it even,” falling for the “adding an even number makes the result even” misconception. But \( n + 2 \) is only even when \( n \) is even. For example, \( n = 3 \) gives 5, which is odd. The parity of the result depends on the parity of \( n \), not on the 2 being added.

Example: 3 and 4 — because \( 9 + 16 = 25 \), which is odd.

Another: 1 and 2 — because \( 1 + 4 = 5 \), which is odd.

Creative: 100 and 7 — because \( 10000 + 49 = 10049 \), which is odd. Any pairing of one even and one odd integer will work.

Trap: 3 and 5 — a student might think “two odd numbers should give an odd result,” which is the “odd + odd = odd” misconception. But \( 9 + 25 = 34 \), which is even. The sum of two odd squares is always even (odd + odd = even). To get an odd sum of squares, you need one even and one odd number.

Example: 9 — because \( 9 = 4 + 5 \) and \( 9 = 2 + 3 + 4 \).

Another: 15 — because \( 15 = 7 + 8 \) and \( 15 = 4 + 5 + 6 \).

Creative: 3 — because \( 3 = 1 + 2 \) and \( 3 = 0 + 1 + 2 \). Including 0 as one of the consecutive integers is valid and often overlooked.

Trap: 6 — a student might offer this because \( 6 = 1 + 2 + 3 \) (sum of three consecutive integers), but 6 cannot be written as the sum of two consecutive integers. This targets the misconception that any multiple of 3 can also be written as a sum of two consecutive integers. Since \( n + (n + 1) = 2n + 1 \) is always odd and 6 is even, it fails. Only odd multiples of 3 satisfy both conditions.

Example: \( 6n \) — this is 6 times an integer, so always a multiple of 6.

Another: \( 12n + 18 \) — this equals \( 6(2n + 3) \), which has a factor of 6.

Creative: \( n(n + 1)(n + 2) \) — the product of three consecutive integers. Among any three consecutive integers, one must be a multiple of 2 and one must be a multiple of 3, so the product always has factors of both 2 and 3, making it a multiple of 6.

Trap: \( 2n + 3 \) — a student might think “the 2 gives a factor of 2 and the 3 gives a factor of 3, so it’s a multiple of 6.” This is the “coefficients vs factors” misconception — having the digits 2 and 3 appear in an expression does not make it divisible by 6. The expression \( 2n + 3 \) is simply an odd number for any integer \( n \) (e.g., \( n = 1 \) gives 5). For an expression to always be a multiple of 6, 6 (or both 2 and 3) must be factors of the entire expression, not just coefficients or constants within it.

Let the two consecutive integers be \( n \) and \( n + 1 \). Their sum is \( n + (n + 1) = 2n + 1 \). Since \( 2n + 1 \) is one more than an even number, it is always odd — regardless of whether \( n \) itself is odd or even.

Students sometimes think they need to check odd and even cases separately, falling into the “proof by cases when algebra suffices” trap. The algebra shows it doesn’t matter: the +1 ensures the sum is always odd.

Let the even number be \( 2a \) and the odd number be \( 2b + 1 \). Their sum is \( 2a + 2b + 1 = 2(a + b) + 1 \), which is always odd.

This targets the “parity rule confusion” misconception. Students frequently muddle the rules for adding even and odd numbers. The sum of an even and an odd number is always odd, never even. A quick check confirms: \( 4 + 3 = 7 \) (odd), \( 10 + 1 = 11 \) (odd).

This is sometimes true and sometimes false. True case: When \( n = 3 \), \( n^2 = 9 > 3 \). In general, \( n^2 > n \) when \( n > 1 \) or \( n < 0 \). False case: When \( n = 1 \), \( n^2 = 1 \), so \( n^2 = n \), not greater. When \( n = 0 \), \( n^2 = 0 = n \).

This probes the “squaring always makes bigger” misconception. Students who test only values like 2, 3, 4 will conclude “always.” But \( n = 0 \) and \( n = 1 \) are both counterexamples. It is important to also test negative integers: \( (-2)^2 = 4 > -2 \). This is true because squaring a negative number makes it positive, which is always greater than the original negative number.

Let the four consecutive integers be \( n \), \( n + 1 \), \( n + 2 \), and \( n + 3 \). Their sum is \( 4n + 6 = 4(n + 1) + 2 \). This is always 2 more than a multiple of 4, so it is never a multiple of 4.

This targets the “sum of n consecutive integers is always a multiple of n” misconception. The sum of three consecutive integers IS always a multiple of 3, and the sum of five consecutive integers IS always a multiple of 5. Students naturally assume the pattern extends to all values of \( n \), but the algebra shows it fails for \( n = 4 \) (and all other even values of \( n \)).

The student’s conclusion happens to be correct, but their method is not a valid proof. This is the “verification is not proof” misconception — testing specific numbers, no matter how many, cannot prove a statement works for all integers. A proof requires algebra: let the five consecutive integers be \( n, n + 1, n + 2, n + 3, n + 4 \). Their sum is \( 5n + 10 = 5(n + 2) \), which is always a multiple of 5.

The danger of relying on examples is that some patterns do break down. For instance, the sum of four consecutive integers is never a multiple of 4, yet any small set of examples might not make this obvious. Only algebra can prove a result holds universally.

The student reaches the correct conclusion — the sum of any two odd numbers is indeed always even — but their proof contains the “same variable for independent unknowns” error. By writing the two odd numbers as \( 2n + 1 \) and \( 2n + 3 \), the student has only considered consecutive odd numbers (numbers that are exactly 2 apart), not any two odd numbers.

The correct approach uses two independent variables: let the odd numbers be \( 2a + 1 \) and \( 2b + 1 \). Then \( (2a + 1) + (2b + 1) = 2a + 2b + 2 = 2(a + b + 1) \), which is even. This covers all pairs of odd numbers, not just those that are 2 apart. The student’s proof also incorrectly implies the sum is always a multiple of 4, which is false — for example, \( 1 + 5 = 6 \), which is even but not a multiple of 4.

This is the classic “squaring a sum” misconception — the student believes that \( (a + b)^2 = a^2 + b^2 \), missing the crucial middle term. The correct expansion uses the identity \( (a + b)^2 = a^2 + 2ab + b^2 \): \( (n + 4)^2 = n^2 + 2(n)(4) + 4^2 = n^2 + 8n + 16 \).

A quick numerical check shows the error: when \( n = 1 \), \( (1 + 4)^2 = 25 \), but the student’s answer gives \( 1 + 16 = 17 \). The missing \( 8n \) term makes a significant difference.

The student’s algebra is correct — the sum is indeed \( 4n + 2 \) — but their conclusion is wrong. This is the “partial factor” misconception: the student sees \( 4n \) and assumes the entire expression is a multiple of 4, ignoring the \( + 2 \). The expression \( 4n + 2 = 4(n) + 2 \), which is always 2 more than a multiple of 4, so it is never a multiple of 4.

For example, \( 2 + 4 = 6 \) (not a multiple of 4), \( 4 + 6 = 10 \) (not a multiple of 4), \( 6 + 8 = 14 \) (not a multiple of 4). The sum of two consecutive even numbers is always even, but it is never a multiple of 4.