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How to use quaternions express number like 49 as sums of four squares based on their factorizations?

I just start to learn quaternion in my math class, I tired to find my class note :

Quaternions look like a + b i + c j + d k .

Hmmmm ... i guess a,b,c,d are integers.Real numbers commutate i,j,k?

and..sum of four squares will look like this?

$8 = 4+4+0+0 $

$9 = 4+4+1+0 $

$10 = 4+4+1+1$

$49 =？　$

kataham
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  • What exactly do you want ? Just expressing a number as a sum of $4$ squares, or expressing a number $n$ with quaternions given a representation $n=a^2+b^2+c^2+d^2$ with integers $a,b,c,d$ ? – Peter Nov 28 '16 at 20:45
  • Perhaps if you write $q = a+bi+cj+dk$ and multiply it by its conjugate $\overline{q} = a-bi-cj-dk$ you find $q\overline{q} = a^2+b^2+c^2+d^2$. – JSchoone Nov 28 '16 at 20:51
  • $49 = 49+0+0+0$ – Tpofofn Nov 28 '16 at 22:10

1 Answers1

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Quaternions $q=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}$ have a "norm" defined by

$$ |a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}|^2 = a^2+b^2+c^2+d^2. $$

This is a generalization of $|a+bi|=a^2+b^2$ defined for complex numbers. Indeed, it is still multiplicative: we have $|xy|=|x||y|$ for quaternions $x$ and $y$.

How can we use this idea to write whole numbers as sums of four squares? Let's do a more illustrative example first: say $42$. To write $42=a^2+b^2+c^2+d^2$ we want to find a quaternion $a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}$ with square norm $42$. We can factor $42$ with smaller numbers in various ways, for example as $6\cdot 7$ or $3\cdot 14$. So to write $42$ as $|x|^2$, it is sufficient to find $x$s for which $|x|^2$ is $6$ and $7$, or $3$ and $14$. Indeed, we have

$$ 6=0^2+1^2+1^2+2^2 \\ 7=1^2+1^2+1^2+2^2 $$

(one may permute the summands any way one likes) and then

$$ \begin{array}{l} 6=|\mathbf{i}+\mathbf{j}+2\mathbf{k}|^2 \\ 7 =|1+\mathbf{i}+\mathbf{j}+2\mathbf{k}|^2 \end{array}$$

which implies

$$ \begin{array}{ll} 42 & =|(\mathbf{i}+\mathbf{j}+2\mathbf{k})(1+\mathbf{i}+\mathbf{j}+2\mathbf{k})|^2 \\ & =|-6+\mathbf{i}+\mathbf{j}+2\mathbf{k}|^2 \\ & =6^2+1^2+1^2+2^2. \end{array} $$

Alternatively,

$$ \begin{array}{r} 3 = 1^2+1^2+0^2+1^2 \\ 14 = 0^2+1^2+2^2+3^2 \end{array} $$

which tells us

$$ \begin{array}{l} \phantom{1}3 = |1+\mathbf{i}+\mathbf{k}|^2 \\ 14 = |\mathbf{i}+2\mathbf{j}+3\mathbf{k}|^2 \end{array} $$

which implies

$$ \begin{array}{ll} 42 & = |(1+\mathbf{i}+\mathbf{k})(\mathbf{i}+2\mathbf{j}+3\mathbf{k})|^2 \\ & = |-4-\mathbf{i}+5\mathbf{k}|^2 \\ & = 4^2+1^2+0^2+5^2. \end{array} $$

I'll let you play around with $49=7\cdot7$. We can write $7=1^2+1^2+2^2+1^2$, but there are different ways to permute the summands. See what they tell you.

Note that you will need to know how to multiply two quaternions to do this. This involves using the distributive property and the multiplication table for $\mathbf{i},\mathbf{j},\mathbf{k}$ (or a helpful mnemonic).

anon
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