Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
\[13^2=\frac{a^2}{b^2}+\frac{m^2}{n^2}+1\]
\[13^2=\frac{(pm)^2}{(qn)^2}+\frac{m^2}{n^2}+1\]\[13^2=\frac{(pm)^2+(qm)^2}{(qn)^2}+1\]
\[a,b,m,n \in \mathbb{Z}\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Curse you, Diophantus!
:)
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! :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question