By Terence Tao

Authored by way of a number one identify in arithmetic, this enticing and obviously awarded textual content leads the reader throughout the numerous strategies excited by fixing mathematical difficulties on the Mathematical Olympiad point. masking quantity idea, algebra, research, Euclidean geometry, and analytic geometry, fixing Mathematical difficulties contains various routines and version options all through. Assuming just a simple point of arithmetic, the textual content is perfect for college kids of 14 years and above in natural arithmetic.

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Then we are trying to prove something of the form (factor 1) × (factor 2)|(expression). If the two factors are coprime, then our objective is equivalent to proving both of (factor 1)|(expression) and (factor 2)|(expression) separately. This should be simpler to prove: it is easier to prove divisibility if the divisors are smaller. But there is an annoying ‘2’ in the way. * The cases are quite similar and I will only do the case when n is even. In this case we can write n = 2m (so as to avoid staring at messy ‘n/2’ terms in the following equations—little housekeeping things like this help a solution run smoothly).

So by looking at the shape of the question, we have the following objectives to keep in mind: (a) Express the numerator as a mathematical expression, so that we can manipulate it. (b) Aim to reduce the problem from a p2 -divisiblity problem to something simpler, perhaps a p-divisibility problem. Let us tackle (a) first. First of all, we can get a numerator easily, but not the reduced numerator necessarily. By adding up the fractions under a common denominator we get (2 × 3 × · · · × (p − 1) + 1 × 3 × · · · × (p − 1) + · · · + 1 × 2 × 3 × · · · × (p − 2)) .

Perhaps f (f (2)) is at least 2. ) With this line of thought one can apply (10) yet again: f (3) ≥ f (f (2)) + 1 ≥ f (f (2) − 1) + 1 + 1 ≥ 3. Here we plugged f (2) − 1 into the ‘n’ of our formula. This works because we already know that f (2) − 1 is at least 1. So it seems we can deduce that f (n) ≥ n. Because we used the fact that f (2) was at least 2 to prove that f (3) was at least 3, the general proof reeks of induction. The induction is just a little tricky though. Consider the next case, showing that f (4) ≥ 4.

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