• "Prove algebraically": build a fully general argument using algebra — numerical examples score zero, no matter how many you test.
• "Show that": you're given the target expression (often from a geometric context) and must derive it step by step from a stated principle (e.g. Pythagoras, an area formula).
• "Disprove" / "show [x] is wrong": find just one counter-example. No algebra required — strategic number-testing is the actual skill.

Three non-negotiable steps:

1. Premise — define your variables using the dictionary above.
2. Manipulation — expand, simplify, collect like terms.
3. Conclusion — factorise into the exact form the question claims (e.g. $2k$ for "even", $8k$ for "multiple of 8"), then write a sentence explicitly linking this back to the original claim. Stopping at the factorised algebra loses the final mark — you must state the conclusion in words.

Worked example: Prove the sum of two consecutive odd integers is always even.

Let the integers be $2n+1$ and $2n+3$. Their sum is $(2n+1)+(2n+3) = 4n+4 = 2(2n+2)$.

Since $n$ is an integer, $2n+2$ is also an integer, so $2(2n+2)$ is $2\times\text{integer}$ — the definition of an even number. Therefore the sum is always even.

Many "prove a multiple of..." questions become much faster if you spot $a^2-b^2=(a+b)(a-b)$ instead of brute-force expanding.

Worked example: Prove $(3n+1)^2-(3n-1)^2$ is always a multiple of 12.

Let $a=3n+1,\ b=3n-1$. Then $a+b=6n$ and $a-b=2$, so:

This is $12\times$ an integer, so the expression is always a multiple of 12 — found far faster than expanding both brackets separately.

Test boundaries ($n=0,1$) and special cases (the only even prime, $2$; negative numbers) rather than random "easy" numbers.

Worked example: Ali thinks $n^2+n+1$ is always prime. Testing $n=1,2,3$ gives $3,7,13$ — all prime, looking convincing. But $n=4$: $16+4+1=21=3\times7$, not prime. $n=4$ is a counter-example, so Ali is wrong.

1. Identify the geometric principle (Pythagoras, area of a rectangle/trapezium, etc.).
2. Substitute the given algebraic lengths into the formula and set equal to the given value.
3. Expand and rearrange until it exactly matches the target equation — show every line.

Worked example: A right-angled triangle has sides $x$, $(x+2)$, $(x+3)$ cm (hypotenuse $x+3$). Show the equation simplifies to $x^2-2x-5=0$.

By Pythagoras: $x^2+(x+2)^2=(x+3)^2$

If $n$ is an integer greater than 1, prove that $n^3-n$ is a multiple of 6.

Factorise: $n^3-n = n(n^2-1) = n(n-1)(n+1) = (n-1)(n)(n+1)$.

This is the product of three consecutive integers. Among any three consecutive integers, at least one is even (a multiple of 2) and exactly one is a multiple of 3. So the product is divisible by both 2 and 3, hence by $2\times3=6$. Therefore $n^3-n$ is always a multiple of 6.