Question

Show that 168 cannot be expressed as the sum of the squares of two rational numbers.

Answer 1

\[13^2=\frac{a^2}{b^2}+\frac{m^2}{n^2}+1\]

Answer 2

\[13^2=\frac{(pm)^2}{(qn)^2}+\frac{m^2}{n^2}+1\]\[13^2=\frac{(pm)^2+(qm)^2}{(qn)^2}+1\]

Answer 3

\[a,b,m,n \in \mathbb{Z}\]

Answer 4

Curse you, Diophantus!

Answer 5

:)

Answer 6

Here are my thoughts - if:\[168=a^2+b^2\]then either a and b are both even or both odd. take the case of both odd so a=2m+1 and b=2n+1:\[168=4m^2+4m+4n^2+4n+2\]which leads to:\[166=4(m^2+n^2+m+n)\]but 166 is not evenly divisible by 4 so this case can be rejected.
now take both a and b as even, so a=2m and b=2n leads to:\[168=4m^2+4n^2\]thus:\[42=m^2+n^2\]strange how the number 42 appears everywhere :)
I can continue on this train of thought - but do you think it will lead anywhere good?

Answer 7

Are you assuming that a and b are integers?

Answer 8

ah! sorry - didn't read your question properly - let me think again...

Answer 9

Fair enough, I made the same mistake intially.

Answer 10

hmmm - this is beyond my current understanding. However, I did find this article that has a very similar problem - maybe you will be able to understand it better:

https://docs.google.com/viewer?a=v&q=cache:zNNprPwolosJ:www.math.ucsd.edu/~okikiolu/104b/hws4.pdf+&hl=en&gl=uk&pid=bl&srcid=ADGEESjuK9ScKokS94ta6viGZM-sAr6CpRkDKkoZCaHDIYTKGyysQ3HV4V9WXpsfgaE3-1QNepW2Q-3KKfpDfhJpOfRb3e8q0wGZTkkcD9AfH-IYVEtu7DqwdSfxSG_443AnysJK3vDs&sig=AHIEtbQavqEWSt-SGk2LHnfyVAiU8D1IHA

It is on the first page - problem 4: Show that 21 cannot be expressed as the sum of squares of two rational numbers.

Answer 11

This is from the very interesting http://www.komal.hu/verseny/feladat.cgi?a=honap&h=201211&t=mat&l=en
if anyone's interested.

@asnaseer , thanks for the link

Answer 12

thanks @henpen - gives me more things to learn! :)