Sum of three squares of rationals (proof)

• Apr 11th 2009, 02:34 AM
pinkparrot
Sum of three squares of rationals (proof)
Hi, I'd be grateful if someone could help me prove that 'if a positive rational is the sum of three squares of rational numbers, then its reciprocal is the sum of three squares of rational numbers too'. Thanks in advance :)
• Apr 11th 2009, 02:53 AM
bobak
Maybe i am missing the point here, but i don't see how the sum of three squares parts is relevant at all, the reciprocal of a rational is always rational.

maybe you teacher just want to see how to can handle a bit algebra.

$\displaystyle \frac{1}{\frac{a^2}{b^2} + \frac{c^2}{d^2} + \frac{e^2}{f^2}}$

can you multiply that by $\displaystyle (bdf)^2$ ?

Bobak
• Apr 11th 2009, 03:02 AM
pinkparrot
Quote:

Originally Posted by bobak
Maybe i am missing the point here, but i don't see how the sum of three squares parts is relevant at all, the reciprocal of a rational is always rational.

maybe you teacher just want to see how to can handle a bit algebra.

$\displaystyle \frac{1}{\frac{a^2}{b^2} + \frac{c^2}{d^2} + \frac{e^2}{f^2}}$

can you multiply that by $\displaystyle (bdf)^2$ ?

Bobak

Well, the reciprocal of a rational must be a rational too, that's obvious, but it's not the point. The problem is, how to show that
if: $\displaystyle \frac{p}{q}=\frac{a^2}{b^2} + \frac{c^2}{d^2} + \frac{e^2}{f^2}$
then: $\displaystyle \frac{q}{p}=\frac{k^2}{l^2} + \frac{m^2}{n^2} + \frac{r^2}{s^2}$, so when a positive rational is the sum of three squares of rational numbers, then its reciprocal is also the sum of three squares of rational numbers. Anyway, thanks for trying to help :)
• Apr 11th 2009, 04:15 AM
Soroban
Hello, pinkparrot!

I was skeptical so I cranked out some examples:

. . $\displaystyle \left(\frac{1}{2}\right)^2 + \left(\frac{1}{3}\right)^2 + \left(\frac{1}{6}\right)^2 \;=\;\frac{7}{18} \quad\Longrightarrow\quad \frac{18}{7} \;=\;\left(\frac{1}{7}\right)^2 + \left(\frac{2}{7}\right)^2 + \left(\frac{11}{7}\right)^2$

. . $\displaystyle \left(\frac{1}{2}\right)^2 + \left(\frac{1}{4}\right)^2 + \left(\frac{1}{8}\right)^2 \;=\;\frac{21}{64} \quad\Longrightarrow\quad \frac{64}{21} \;=\;\left(\frac{8}{21}\right)^2 + \left(\frac{16}{21}\right)^2 + \left(\frac{32}{21}\right)^2$

. . $\displaystyle \left(\frac{1}{2}\right)^2 + \left(\frac{1}{3}\right)^2 + \left(\frac{1}{4}\right)^2 \;=\;\frac{61}{144} \quad\Longrightarrow\quad \frac{144}{61} \;=\;\left(\frac{8}{61}\right)^2 + \left(\frac{16}{61}\right)^2 + \left(\frac{92}{61}\right)^2$

. . $\displaystyle \left(\frac{1}{2}\right)^2 + \left(\frac{2}{3}\right)^2 + \left(\frac{3}{4}\right)^2 \;=\;\frac{181}{144} \quad\Longrightarrow\quad \frac{144}{181} \;=\;\left(\frac{8}{181}\right)^2 + \left(\frac{20}{181}\right)^2 + \left(\frac{160}{181}\right)^2$

Along the way I picked up a few ideas, but I don't see a usable pattern.

Do you?

Edit: corrected my third example.
. . . .Thanks for the heads-up, MM.
.
• Apr 11th 2009, 04:36 AM
pinkparrot
Wouldn't be asking for help if I did ;) Your examples show that if $\displaystyle \frac{q}{p}$ is the reciprocal, the denominators of the 3 square elements seem to equal $\displaystyle p^2$. But even assuming it's always so doesn't get me any further...
• Apr 11th 2009, 07:34 AM
Media_Man
A constructive proof...
This is actually quite trivial. The trick is to look at the rationals before they are reduced:

Let $\displaystyle p/q = (a/b)^2 + (c/d)^2 + (e/f)^2$ . Now let $\displaystyle L=lcm(b,d,f)$. Rewrite as $\displaystyle p/q = P/L^2 = (A/L)^2 + (C/L)^2 + (E/L)^2 = (A^2 + C^2 +E^2)/L^2$, so $\displaystyle P=(A^2 + C^2 +E^2)$ .

Then $\displaystyle (AL/P)^2 + (CL/P)^2 +(EL/P)^2 = (L^2/P^2)(A^2 + C^2 +E^2) = (L^2/P^2)P = L^2/P = q/p$

Now simply reduce $\displaystyle AL/P = k/l, CL/P = m/n, EL/P = r/s$ . This is a constructive proof yielding a “natural” answer. The fact that other answers exist is only icing on the cake.

Example: $\displaystyle (1/2)^2 + (1/3)^2 + (1/6)^2 = 7/18$ can be rewritten as $\displaystyle (3/6)^2 + (2/6)^2 + (1/6)^2 = 14/36$. Therefore $\displaystyle (3*6/14)^2 + (2*6/14)^2 + (1*6/14)^2 = (6/14)^2 *(3^2 + 2^2 + 1^2) = (6/14)^2 * 14 = 6^2/14$. So the “natural” solution is $\displaystyle 18/7 = (9/7)^2 + (6/7)^2 + (3/7)^2$ .
Likewise, $\displaystyle 144/181 = (72/181)^2 + (96/181)^2 + (108/181)^2$ . Soroban’s second example is already in “natural” form, and the third example has a typo in it: $\displaystyle (1/2)^2 + (1/3)^2 + (1/6)^2$ is NOT $\displaystyle 31/30$
• Apr 12th 2009, 03:58 AM
pinkparrot
Indeed, simple and constructive. Thanks a lot, Media_Man!
• Apr 12th 2009, 08:26 AM
Media_Man
No problem. Interestingly, there is a "four squares theorem" that eluded the likes of Gauss and Euler which states that all positive integers can be represented by the sum of four or fewer squares. Most require only three, the ones that require exactly four being 7, 15, 23, 28, 31, 39... (Sloane's A004215). I wonder if these can be expressed by the sum of three rational, non-integer squares? Even if not, I wonder if the four squares theorem can be extended to rationals as well...