Algebraic Proof

An odd integer is one more than an even integer.

1. 2n is even for every integer n.
2. Adding 1 makes it odd, so 2n+1 is always odd.
3. For instance, n^2+1 is even when n=1, so that option is not always odd.

Add the expressions before collecting like terms.

1. Add the two expressions: 2n+(2n+2).
2. Collect the terms to obtain 4n+2.

A multiple of a number can be written as that number times an integer.

1. To show a number is divisible by 3, write it as 3\times(an integer).
2. Factor out 3: 6n+9=3(2n+3).
3. For integer n, 2n+3 is an integer, so 6n+9 is an integer multiple of 3.
4. The third option has the right factorisation but the wrong reason: 2n+3 need not be a multiple of 3 (when n=1 it is 5). It only needs to be an integer.

Try each pair with an even value of n and an odd value of n.

1. 2n is always even, so 2n+1 is always odd.
2. Consecutive odd numbers differ by 2, so the next odd number is 2n+3.
3. 2n+2 is even, and 2n and 2n+2 are both even.
4. n+1 and n+3 are odd only when n is even: n=1 gives 2 and 4. So they do not always work.

Use the parity of two consecutive integers.

1. Write the expression as n(n+1)+5.
2. n and n+1 are consecutive integers, so one of them is even, and their product n(n+1) is even.
3. An even number plus the odd number 5 is odd, so n^2+n+5 is always odd.
4. The first option is false: n(n+1) is never odd, because one of the two factors is always even.

Expand the square, then interpret the simplified expression.

Part (a)

1. Expand (n+1)^2 as n^2+2n+1.
2. Subtracting n^2 leaves 2n+1.

Part (b)

1. The simplified difference is 2n+1.
2. Since 2n is even, the difference is one more than an even integer and is therefore odd.
3. The first option is wrong: twice an integer is always even.

Expand the product and group the terms that are multiples of four.

Part (a)

1. Multiply each term in the first bracket by each term in the second.
2. (2n+1)(2n+3)=4n^2+6n+2n+3.
3. The simplified product is 4n^2+8n+3.

Part (b)

1. The first two terms together are 4n(n+2), a multiple of 4.
2. The product is 3 more than that multiple, so its remainder is 3.

A counterexample needs only one value that makes the claim false.

1. At n=10, the expression is 100+10+11=121.
2. Since 121=11^2, it is not prime.
3. One counterexample is enough to disprove a claim made for every positive integer.

Remember that subtracting the second square changes every sign in its expansion.

1. Expand the first square: 9n^2+12n+4.
2. The second square is 9n^2-12n+4.
3. Subtract the whole second expression; the squared and constant terms cancel.
4. The result is 24n.

A difference of two squares can be factorised without expanding both squares.

Part (a)

1. Use the difference of squares: E=((n+4)-(n-2))((n+4)+(n-2)).
2. This gives 6(2n+2).
3. Factor out the remaining 2 to get 12(n+1).

Part (b)

1. The factorised expression is 12(n+1), where n+1 is an integer.
2. Every integer multiple of 12 is divisible by both 3 and 4.
3. It need not be divisible by 24; take n=0.

A proof of a universal claim must cover every allowed value.

1. Each check establishes the claim only for that particular value.
2. There are infinitely many integers, including values outside those five.
3. A general argument is needed, such as writing the expression as the product of two consecutive integers.

Complete the square to find a part that is never negative.

1. Complete the square: x^2+4x=(x+2)^2-4, so x^2+4x+7=(x+2)^2-4+7=(x+2)^2+3.
2. A square is never negative, so (x+2)^2\ge0 and the expression is at least 3.
3. Since 3>0, the expression is positive for all values of x.
4. The first option is wrong: (x+2)^2-3 expands to x^2+4x+1, not x^2+4x+7.

Translate the whole trick into an algebraic expression before simplifying.

Part (a)

1. The final result is (n+3)^2-n^2-9.
2. Expand the square: n^2+6n+9-n^2-9.
3. The result simplifies to 6n.

Part (b)

1. The simplified result is 6n.
2. Because n is an integer, this is an integer multiple of 6, which proves the claim.

Expand each square carefully; the n terms cancel.

Part (a)

1. Expand each square: (n-1)^2=n^2-2n+1 and (n+1)^2=n^2+2n+1.
2. Add the three squares: n^2-2n+1+n^2+n^2+2n+1.
3. The n terms cancel, leaving 3n^2+2.

Part (b)

1. 3n^2 is a multiple of 3, because n^2 is an integer.
2. So 3n^2+2 is always 2 more than a multiple of 3: the remainder is 2, whichever integers are used.
3. Check: 0^2+1^2+2^2=5, which is 3+2.

Factorise the numerator and use the fact that the denominator is nonzero.

Part (a)

1. The numerator factorises as (n+1)(n+2).
2. Since n is positive, n+1 is nonzero and can be cancelled.
3. The quotient is n+2.

Part (b)

1. The simplified quotient is n+2.
2. Adding the integer 2 to the integer n gives an integer.
3. This establishes the claim for every allowed n.

Expand the squared bracket and factorise the result completely.

Part (a)

1. Expand the square: (2n+1)^2=4n^2+4n+1.
2. Subtract 9: 4n^2+4n-8.
3. Take out the common factor 4: 4(n^2+n-2).
4. Factorise the quadratic: n^2+n-2=(n-1)(n+2), so the answer is 4(n-1)(n+2).
5. (Or use the difference of two squares: (2n+1-3)(2n+1+3)=(2n-2)(2n+4)=4(n-1)(n+2).)

Part (b)

1. n-1 and n+2 differ by 3, which is odd, so one of them is even and the other is odd.
2. So (n-1)(n+2) is even: it is 2\times(an integer).
3. Then 4(n-1)(n+2)=4\times2\times(an integer) =8\times(an integer), a multiple of 8.
4. The first option is false: 12 is a multiple of 4 but not of 8.

Expand and look for a perfect-square trinomial.

Part (a)

1. Expand the product: 4n^2+4n-3.
2. Adding 4 gives 4n^2+4n+1.
3. This is the perfect square (2n+1)^2.

Part (b)

1. The expression is (2n+1)^2.
2. For integer n, the number 2n+1 is odd.
3. Squaring it gives an odd square number.

First take out the common factor and then factorise the remaining quadratic.

Part (a)

1. Take out the common factor n: n(n^2+3n+2).
2. Factorise the quadratic: n^2+3n+2=(n+1)(n+2).
3. The fully factorised expression is n(n+1)(n+2).

Part (b)

1. n, n+1 and n+2 are three consecutive integers.
2. At least one of them is even, and exactly one of them is a multiple of 3.
3. So their product has factors 2 and 3, which makes it divisible by 6.
4. The first option is false: 1\times1\times1=1 is not divisible by 6. The integers must be consecutive.

Odd numbers go up in steps of 2. After expanding, split the constant so that you can take out a factor of 8.

Part (a)

1. The next odd number is 2 more than 2n+1, so it is 2n+3.
2. Expand both squares: (2n+1)^2=4n^2+4n+1 and (2n+3)^2=4n^2+12n+9.
3. Add them: 8n^2+16n+10.

Part (b)

1. Split the constant: 8n^2+16n+10=8n^2+16n+8+2=8(n^2+2n+1)+2.
2. 8(n^2+2n+1) is a multiple of 8, because n^2+2n+1 is an integer.
3. So the sum is always 2 more than a multiple of 8: the remainder is 2.
4. Check: 1^2+3^2=10=8+2, and 3^2+5^2=34=32+2.

Expand the cube and factorise what remains.

Part (a)

1. Expand the cube: (n+1)^3=(n+1)(n^2+2n+1)=n^3+3n^2+3n+1.
2. Subtract n^3 and 1: 3n^2+3n.
3. Take out the common factor 3n: 3n(n+1).

Part (b)

1. n and n+1 are consecutive integers, so one of them is even and n(n+1) is even.
2. So 3n(n+1)=3\times2\times(an integer) =6\times(an integer), a multiple of 6.
3. The third option is false: 9 is a multiple of 3 but not of 6.