Mathematical Food for Thought

I’ll be leaving tomorrow morning for Boston, Massachusetts! Probably no blog posts for 6 weeks! Take a break from math! Or not!

Problem: Find a closed form for the sum

.

Solution: Consider the expansion of

.

It’s not hard to see that if we let the desired sum be , we have

since each pair is counted twice and the squared terms are there. But we can simplify the sums by well-known formulas to get

,

which becomes

after massive simplification. QED.

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Comment: Symmetry was definitely the best way to approach this problem (or as far as I know anyway). You could’ve found a lot of ugly summations to get to the desired expression as well.

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Practice Problem: Can you generalize? In any way, shape, or form.

Problem: (1991 India National Olympiad – #10) For any positive integer , let denote the number of ordered pairs of positive integers for which . Determine the set of positive integers for which .

Solution: Rewrite the equation as

.

So for each pair with , there exist a pair satisfying

.

Hence we want to find an with distinct factors. We see that for prime is the only case (since itself is prime, meaning is not possible), so we thus have for , where is any prime. QED.

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Comment: A standard problem involving Simon’s Favorite Factoring Trick and counting the number of divisors. SFFT is always something to look for when you have terms (sorta like completing the square except not).

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Practice Problem: (1991 India National Olympiad – #1) Find the number of positive integers for which

(i) ;

(ii) is a factor of .

Problem: (1991 India National Olympiad – #7) Solve the following system for real

.

Solution: Well, we simply go about solving this system the regular way – look for something to eliminate. We try using the first and second equations and find

,

which we subtract from the second equation to get

.

Using the third equation, we have so we substitute, multiply through by (it can’t be zero), and divide by to get

.

So . But from the second equation, we see that gives but they are real so this is impossible. Hence . It remains to solve for and using

.

Substitute into the first one and again multiply through by to find

.

We then have with corresponding . Checking, we see that both solutions work, so our final solution set is

; .

QED.

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Comment: Not bad for an olympiad question – just requires basic algebraic manipulation and solving quadratics. Looking for the right things to square and substitute made this problem considerably easier.

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Practice Problem: (1994 India National Olympiad – #2) If , prove that .