General algebraic proofs

(a) $2n + (2n + 2) + (2n + 4) = 6n + 6 = 6(n + 1)$

1. Write three consecutive even numbers in terms of $n$.
   $$2n, \quad 2n + 2, \quad 2n + 4$$
2. Add.
   $$2n + (2n + 2) + (2n + 4) = 6n + 6$$
3. Factorise.
   $$6n + 6 = 6(n + 1)$$
4. $n + 1$ is an integer, so the sum is a multiple of 6.

• M1 Writing consecutive even numbers as $2n$, $2n + 2$, $2n + 4$.
• M1 Simplifying to $6n + 6$.
• C1 Writing $6(n + 1)$ and concluding that it is a multiple of 6.

(a) When $n = 4$, $16 + 4 + 1 = 21 = 3 \times 7$, which is not prime.

1. Try values of $n$: $n = 1$ gives 3, $n = 2$ gives 7, $n = 3$ gives 13, $n = 4$ gives 21.
2. $21 = 3 \times 7$ is not prime, so Kai is wrong.

• M1 Testing values until a non-prime result, such as $n = 4$ giving 21.
• C1 Stating that 21 (or another value) is not prime, with its factors, so one counterexample shows Kai is wrong.

(b) $n^2 + n = n(n + 1)$, a product of consecutive integers, one of which is even.

1. Factorise.
   $$n^2 + n = n(n + 1)$$
2. $n$ and $n + 1$ are consecutive integers, so one of them is even. A product with an even factor is even.

• M1 Factorising to $n(n + 1)$.
• C1 Explaining that one of two consecutive integers is even, so the product is even.

(a) The difference simplifies to $8n$. Since n is an integer this is a multiple of 8.

1. Expand both squares with their signed middle terms.
   $$n^{2}+4n+4-(n^{2}-4n+4)$$
2. Cancel the square and constant terms.
   $$8n$$
3. The difference simplifies to $8n$. Since n is an integer this is a multiple of 8.

• M1 Expand both squares with their signed middle terms.
• M1 Cancel the square and constant terms.
• C1 Correct conclusion with supporting reasoning: The difference simplifies to $8n$. Since n is an integer this is a multiple of 8.

(b) An integer multiplier establishes divisibility. A non-integer multiplier need not produce an integer multiple.

1. An integer multiplier establishes divisibility. A non-integer multiplier need not produce an integer multiple.

• C1 Correct conclusion with supporting reasoning: An integer multiplier establishes divisibility. A non-integer multiplier need not produce an integer multiple.

(a) $(2n+1)(2n+3)=4(n+1)^2-1$, one less than four times the integer $(n+1)^2$.

1. Represent consecutive odd integers by 2n+1 and 2n+3.
   $$(2n+1)(2n+3)$$
2. Expand and group the result around a multiple of 4.
   $$4n^{2}+8n+3=4(n+1)^{2}-1$$
3. Therefore $(2n+1)(2n+3)=4(n+1)^2-1$, one less than four times the integer $(n+1)^2$.

• M1 Represent consecutive odd integers by 2n+1 and 2n+3.
• M1 Expand and group the result around a multiple of 4.
• C1 Correct conclusion with supporting reasoning: $(2n+1)(2n+3)=4(n+1)^2-1$, one less than four times the integer $(n+1)^2$.

(b) No. For example 3 times 5 equals 15, which has factors other than 1 and itself.

1. No. For example 3 times 5 equals 15, which has factors other than 1 and itself.

• C1 Correct conclusion with supporting reasoning: No. For example 3 times 5 equals 15, which has factors other than 1 and itself.

(a) Writing the integers as n-1, n and n+1 gives sum 3n, a multiple of 3.

1. Represent the consecutive integers around their middle integer.
   $$(n-1)+n+(n+1)$$
2. Collect terms into an integer multiple.
   $$3n$$
3. Writing the integers as n-1, n and n+1 gives sum 3n, a multiple of 3.

• M1 Represent the consecutive integers around their middle integer.
• M1 Collect terms into an integer multiple.
• C1 Correct conclusion with supporting reasoning: Writing the integers as n-1, n and n+1 gives sum 3n, a multiple of 3.

(b) 2+3+4=9, which is not divisible by 6.

1. Therefore 2+3+4=9, which is not divisible by 6.

• C1 Correct conclusion with supporting reasoning: 2+3+4=9, which is not divisible by 6.

(a) n² + (n + 1)² = 2n² + 2n + 1 = 2(n² + n) + 1, which is one more than an even number, so it is odd.

1. Expand and collect the sum of the two squares.
   $$n^{2}+(n+1)^{2}=2n^{2}+2n+1$$
2. Therefore n² + (n + 1)² = 2n² + 2n + 1 = 2(n² + n) + 1, which is one more than an even number, so it is odd.

• M1 Expand and collect the sum of the two squares.
• C1 Correct conclusion with supporting reasoning: n² + (n + 1)² = 2n² + 2n + 1 = 2(n² + n) + 1, which is one more than an even number, so it is odd.

(b) 1000 is even, but every such sum is odd.

1. Therefore 1000 is even, but every such sum is odd.

• C1 Correct conclusion with supporting reasoning: 1000 is even, but every such sum is odd.

(a) $(2n - 1)(2n + 1) = 4n^2 - 1$

1. Two consecutive odd numbers.
   $$2n - 1, \quad 2n + 1$$
2. Multiply.
   $$(2n - 1)(2n + 1) = 4n^2 - 1$$
3. $4n^2$ is a multiple of 4, so the product is one less than a multiple of 4.

• P1 Representing consecutive odd numbers as $2n - 1$ and $2n + 1$ (or $2n + 1$ and $2n + 3$).
• M1 Expanding to $4n^2 - 1$ (or $4n^2 + 8n + 3$).
• C1 Writing it as $4(\ldots) - 1$ and concluding.

(b) $(n + 1)^2 - n^2 = 2n + 1$

1. $$(n + 1)^2 - n^2 = n^2 + 2n + 1 - n^2 = 2n + 1$$
2. $2n$ is even, so $2n + 1$ is odd.

• M1 Reaching $2n + 1$.
• C1 Explaining that $2n + 1$ is one more than an even number, so it is odd.

(a) For n = 2k + 1, $n^{2}$ $-$ 1 = 4k(k + 1) is a multiple of 8. Also $n^{2}$ + 1 is even. Hence $n^{4}$ $-$ 1 = ($n^{2}$ $-$ 1)($n^{2}$ + 1) is a multiple of 16.

1. $$(2k+1)^{2}-1=4k(k+1)$$
2. $$n^{4}-1=(n^{2}-1)(n^{2}+1)$$
3. Write n = 2k + 1. Then $n^{2}$ $-$ 1 = 4k(k + 1).
4. Since k and k + 1 are consecutive integers, their product is even. Therefore $n^{2}$ $-$ 1 = 8m for some integer m.
5. An odd integer has an odd square, so $n^{2}$ + 1 = 2r for some integer r.
6. Use the difference of squares: $n^{4}$ $-$ 1 = ($n^{2}$ $-$ 1)($n^{2}$ + 1) = 8m $\times$ 2r = 16mr.

• P1 Establishing $(2k+1)^{2}-1=4k(k+1)$ or an equivalent valid method.
• P1 Establishing $n^{4}-1=(n^{2}-1)(n^{2}+1)$ or an equivalent valid method.
• C1 Correct conclusion with the complete supporting argument: For n = 2k + 1, $n^{2}$ $-$ 1 = 4k(k + 1) is a multiple of 8. Also $n^{2}$ + 1 is even. Hence $n^{4}$ $-$ 1 = ($n^{2}$ $-$ 1)($n^{2}$ + 1) is a multiple of 16.