Writing Math Solutions

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How to Write a Solution
Solution Readers, not Mindreaders
by Richard Rusczyk

A full solution does not just mean a correct answer. You should justify every notable step of your solution. An experienced reader should never wonder 'Why is that true?' while reading your solution. She should also never be left in doubt as to whether or not you know why it is true.

It's not always clear what steps you can assume the reader understands and what steps you have to explain. Here area few guidelines:

If you can cite a theorem that has a name, then you don't have to prove the theorem. You can cite the theorem and move on, as in, 'By the Pythagorean Theorem, AC = 3.'
If you are very confident the step is well known but you don't know a name, you can say 'By a well-known theorem, the area of ABC equals rs, where r is the inradius and s is the semiperimeter.' You can also leave out the 'By a well-known theorem' bit, particularly for extremely common results such as the one just stated. (If you don't know that result, try proving it on your own.)
If you still aren't sure whether to prove a certain step or assume it's well-known, you have a decision to make. If you can prove it in one or two lines, go ahead and do so. If it's going to take a lot of work to prove but you know how to do it, then at least outline the proof (and give a more thorough one if you have time). If you're taking a contest and have no idea how to prove it, cite it and move on. Maybe you'll get lucky and it will be a 'well-known theorem'. If you're writing an educational paper and you don't know how to prove it, then your paper isn't finished until you figure it out.
When writing a string of algebraic steps, each step should follow obviously from the one before it. Don't write something like, 'Thus, we have

x²(x-4) + (x + 1)² + 5x - 4(x + 2) = -2,

so x = 1 is the only solution.' You should include clear simple steps that make it clear that the above is equivalent to (x-1)³ = 0.
You can invoke symmetry or analogy when the cases are precisely the same. For example, suppose you want to prove that the area of any triangle ABC is given by
[ABC] = (ab sin C + bc sin A + ac sin B)/6,
where a = BC, b = AC, and c = AB, and [ABC] is the area of ABC. If you prove that [ABC] = (ab/2)(sin C), then you can just write, 'Similarly, we have
[ABC] = (bc/2)(sin A) = (ac/2)(sin B).'
When in doubt, explain it. Many of the solutions presented in this article have a little overkill in them. It's better to prove too much than too little.

Here is a sample problem:

Problem: Find all integral solutions to the equation

(m²+ 1)(n²+ 1) + 2(m - n)(1 - mn) = 4(mn + 1).

(Problem by Titu Andreescu)

This is totally unacceptable:

How Not to Write the Solution 1:

(1, 2), (-3, 0), (0, 3), (-2, -1), (1, 0), (-3, 2), (0, -1), (-2, 3)

The above is an answer, not a solution. This 'solution' lacks any evidence that these solutions actually work, and doesn't show that there are no other solutions. Moreover, it brings the reader no closer to understanding the solution.

How Not to Write the Solution:

The given equation rearranges to

(m + 1)(n - 1) = ± 2,

so the solutions are (1, 2), (-3, 0), (0, 3), (-2, -1), (1, 0), (-3, 2), (0, -1), (-2, 3).

The above solution is better than the first one; a motivated reader at least has a glimmer of a path to the solution, but it's not at all clear how the original equation rearranges to the given equation, nor how the show solutions follow.

How to Write the Solution:

We expand the first term and the right-hand side then regroup terms:

(m²+ 1)(n²+ 1) + 2(m - n)(1 - mn) = 4(mn + 1)
m²n²+ m² + n²+ 1 + 2(m - n)(1 - mn) = 4mn + 4
m²n²- 2mn + 1 + m² - 2mn + n² + 2(m - n)(1 - mn) = 4
(mn - 1)² + (m - n)² - 2(m - n)(mn - 1) = 4

This left side is the square of [(mn - 1) - (m - n)], so we have:

[(mn - 1) - (m - n)]² = 4
[mn - m + n - 1]² = 4
[(m + 1)(n - 1)]² = 4
(m + 1)(n - 1) = ±2,

Case 1: For (m + 1)(n - 1) = 2, we have the systems of equations:

m + 1 = 1 n - 1 = 2	m + 1 = 2 n - 1 = 1	m + 1 = -1 n - 1 = -2	m + 1 = -2 n - 1 = -1
(0, 3)	(1, 2)	(-2, -1)	(-3, 0)

Case 2: For (m + 1)(n - 1) = -2, we have the systems of equations:

m + 1 = 1 n - 1 = -2	m + 1 = -2 n - 1 = 1	m + 1 = -1 n - 1 = 2	m + 1 = 2 n - 1 = -1
(0, -1)	(-3, 2)	(-2, 3)	(1, 0)

Thus, the solutions are (1, 2), (-3, 0), (0, 3), (-2, -1), (1, 0), (-3, 2), (0, -1), (-2, 3).

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