Cuboid computation

You are given a ruler which can be extended indefinitely, and an unlimited supply of 1-inch sticks of negligible diameter. They may be glued together to form a cuboid.

If you construct a cuboid, you can measure the length from one vertex to another, and in doing so "compute" the square root of a certain number by measuring a particular diagonal.

What is the natural density (or asymptotic density) of positive integers whose square roots cannot be "computed" in this way?


The answer is 0.166666.

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1 solution

Isaac Buckley
Dec 16, 2015

The trick is to notice the question is really asking what natural numbers can't be writen as the sum of three squares.

Great, luckily for me Legendre worked this one out already. See Legendre's three-square theorem .

It states a number can't be writen as the sum of 3 sqaures iff it's of the form 4 a ( 8 b + 7 ) 4^a(8b+7) .

We can see from this that the density should be 1 8 n = 0 1 4 n = 1 6 \frac{1}{8}\sum\limits_{n=0}^{\infty}\frac{1}{4^n}=\frac{1}{6}

You forgot to include the case for the density of numbers that can't be written as a sum of 2 squares.

Julian Poon - 5 years, 6 months ago

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I'm afraid I'm not sure what you mean.

If it can't be written as the sum of 3 squares it certainly cant be written as the sum of 2 squares, 0 is counted as a square in this context.

Isaac Buckley - 5 years, 6 months ago

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Oh, 0 is counted, I forgot. Ok goodbye :P

Julian Poon - 5 years, 6 months ago

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@Julian Poon Face diagonal still counts.

Jake Lai - 5 years, 5 months ago

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@Jake Lai Even if 0 is not counted, the answer would still be 1 6 \frac{1}{6} either ways.

Julian Poon - 5 years, 5 months ago

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