Loading [MathJax]/extensions/TeX/newcommand.js
Skip to main content

Section 13.5 All the Squares Fit to be Summed

There is one loose end. What are all the numbers we can represent as a sum of squares?

For instance, why are some composite numbers of the form 4k+1 not writeable as the sum of two squares? Also, many even numbers are representable – how do we tell which even numbers are writeable? We conclude our discussion by proving the full statement, after a couple of preliminary lemmas.

Each of those primes is representable, so we can use Fact 13.1.8 to write all intermediate products as a sum of squares. Hence all such products are representable.

Example 13.5.2.

Consider this:

\begin{equation*} 442 =2\cdot 13\cdot 17= \left(1^2+1^2\right)\left( 3^2+2^2\right)\cdot 17 \end{equation*}
\begin{equation*} = \left[\left(1\cdot 3-1\cdot 2\right)^2+\left(1\cdot 2+1\cdot 3\right)^2\right]\left(4^2+1^2\right) \end{equation*}
\begin{equation*} =\left(1^2+5^2\right)\left(4^2+1^2\right)=\left(1\cdot 4-5\cdot 1\right)^2+\left(1\cdot 1+5\cdot 4\right)^2=1^2+21^2\text{.} \end{equation*}

First, \(p^2\) (even if \(p\) is not prime) is trivially always representable, since \(p^2=p^2+0^2\text{.}\) Now, rather than using Fact 13.1.8, let \(P\) be the product of all prime factors of the form \(4k+3\text{,}\) which is necessarily a perfect square \(P=Q^2\text{,}\) given that all the powers are even. We can simply multiply this by \(N/Q^2=a^2+b^2\text{,}\) which is possible by Lemma 13.5.1 since \(Q^2\) removes all primes of the form \(4k+3\) in the prime factorization. This yields \((aQ)^2+(bQ)^2\text{.}\)

Example 13.5.4.

Consider this:

\begin{equation*} 35802=442\cdot 3^4 = \left(1^2+21^2\right)3^2\cdot 3^2 \end{equation*}
\begin{equation*} =1^2\cdot 3^2\cdot 3^2+21^2\cdot 3^2\cdot 3^2=9^2+189^2 \end{equation*}

From 13.5.1 and 13.5.3, the only case left to consider if \(N\) has a prime of the form \(p=4k+3\text{,}\) but to an odd power. This seemed to be the bottleneck in our exploration.

By way of contradiction, suppose that it is possible to write

\begin{equation*} N=a^2+b^2\text{.} \end{equation*}

First, divide this equation by any factors of \(p\) common to \(N\text{,}\) \(a\text{,}\) and \(b\) to get

\begin{equation*} M=c^2+d^2 \end{equation*}

The power of \(p\) we divided by (so that \(N=Mp^\ell\)) must be an even power, since each term on the right-hand side is a perfect square and can only contribute even powers of primes by the Fundamental Theorem of Arithmetic.

Since \(N\) had an odd power of \(p\text{,}\) we know \(M\) still has an odd power of \(p\) dividing it, yet \(p\nmid c,d\text{.}\)

Take everything modulo \(p\) to get the congruence

\begin{equation*} 0\equiv c^2+d^2\text{ (mod }p)\text{.} \end{equation*}

Since \(p \nmid c\text{,}\) we can multiply this congruence by \(\left(c^{-1}\right)^2\) to get

\begin{equation*} 0\equiv 1+\left(c^{-1}\right)^2d^2\Rightarrow -1\equiv \left(c^{-1}d\right)^2\text{ (mod }p) \end{equation*}

This is a contradiction, as by Fact 13.3.2 there is no square root of \(-1\) modulo \(p\) for \(p=4k+3\text{,}\) finishing the proof!

Example 13.5.6.

This theorem fully explains why 21=7\cdot 3 and the others mentioned in Fact 13.1.2 cannot be written as a sum of squares.

If the whole theorem still seems too neat and dried, it can be instructive to get insight by plugging in different n below. When do you get an error, when not?

(As a bonus, can you turn this into an interactive cell? See Sage note 12.6.8.)