STEP algebra and inequalities: guide and practice problems
STEP algebra is rarely hard because of new facts. It is hard because the question hands you an identity or an inequality and expects you to see why it is useful three parts later.
Key ideas
- Factorise before you expand: a difference of two squares hidden by adding and subtracting a term is a STEP favourite.
- Symmetric expressions in a, b, c are controlled by a + b + c, ab + bc + ca and abc, which are the coefficients of a cubic.
- To prove an inequality, try to write the difference as a sum of squares, or use AM-GM on carefully chosen terms.
- Square roots of the form sqrt(A ± 2 sqrt B) often simplify to |sqrt p ± sqrt q|: keep the modulus.
Common traps
- Squaring both sides of an inequality without checking signs.
- Dropping the modulus when taking a square root of a square.
- Proving a statement only for the example values the question used.
Practice problems
Original problems in the style of STEP. Try each one for at least 20 minutes before you open a hint; write a full solution, then compare.
Problem 1: A factorisation that telescopes
- Show that \(x^4+4=(x^2+2x+2)(x^2-2x+2)\).
- Hence show that \(n^4+4\) is prime for exactly one positive integer \(n\), and state it.
- By writing \(\dfrac{n}{n^4+4}\) as the difference of two simpler fractions, find \(\displaystyle S_N=\sum_{n=1}^{N}\frac{n}{n^4+4}\) in terms of \(N\), and deduce the value of \(\displaystyle\sum_{n=1}^{\infty}\frac{n}{n^4+4}\).
Hint 1
For (i), add and subtract \(4x^2\): \(x^4+4=(x^2+2)^2-(2x)^2\).
Hint 2
Note that \(x^2+2x+2=(x+1)^2+1\) and \(x^2-2x+2=(x-1)^2+1\). The difference of their reciprocals has numerator \(4x\).
Full solution
(i) \(x^4+4=(x^2+2)^2-4x^2=(x^2+2-2x)(x^2+2+2x)\).
(ii) For a positive integer \(n\), both factors are integers and \(n^2+2n+2\ge 5\). So \(n^4+4\) can only be prime if \(n^2-2n+2=(n-1)^2+1=1\), that is \(n=1\), and then \(n^4+4=5\), which is prime. So \(n=1\) is the only one.
(iii) \(\dfrac{1}{(n-1)^2+1}-\dfrac{1}{(n+1)^2+1}=\dfrac{(n+1)^2-(n-1)^2}{n^4+4}=\dfrac{4n}{n^4+4}\). So with \(a_m=\dfrac1{m^2+1}\),
\[S_N=\tfrac14\sum_{n=1}^N\left(a_{n-1}-a_{n+1}\right)=\tfrac14\left(a_0+a_1-a_N-a_{N+1}\right)=\frac14\left(\frac32-\frac1{N^2+1}-\frac1{(N+1)^2+1}\right).\]As \(N\to\infty\) the last two terms tend to 0, so the sum is \(\tfrac14\cdot\tfrac32=\tfrac38\).
Results: \(n=1\); \(S_N=\frac14\left(\frac32-\frac1{N^2+1}-\frac1{(N+1)^2+1}\right)\); the infinite sum is \(\frac38\).
2 more algebra and inequalities problems
Three numbers with a fixed sum and pair-sum; Nested square roots. Each with two hints and a full solution.
Next steps
Related topics: Functions and curve sketching · Sequences, series and induction. Stuck? Read the problem-solving strategies. Ready for the real thing? Official STEP past papers.
STEP is run by OCR. A Level Math Revision is independent: it is not affiliated with or endorsed by OCR, the University of Cambridge or any university. Every problem here is our own, written in the style of STEP; for real papers use OCR's official past papers.