Here's the question you clicked on:
henpen
Show that 168 cannot be expressed as the sum of the squares of two rational numbers.
\[13^2=\frac{a^2}{b^2}+\frac{m^2}{n^2}+1\]
\[13^2=\frac{(pm)^2}{(qn)^2}+\frac{m^2}{n^2}+1\]\[13^2=\frac{(pm)^2+(qm)^2}{(qn)^2}+1\]
\[a,b,m,n \in \mathbb{Z}\]
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?
Are you assuming that a and b are integers?
ah! sorry - didn't read your question properly - let me think again...
Fair enough, I made the same mistake intially.
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.
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
thanks @henpen - gives me more things to learn! :)