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