Hi
I have been scratching my head over this problem.
Find s,t,u and v if s^2 + t^2 + u^2 + v^2 - st - tu -uv - v + 2/5 = 0
Please help!
gentsl
I too have been scratching my head over this problem.
Are you using parenthesis correctly?
It mostly looks like a candidate for the following manipulations:
$\displaystyle 2s^2 + 2t^2 + 2u^2 + 2v^2 - 2st - 2tu -2uv - 2v + \frac45 = 0$
$\displaystyle s^2 + (s-t)^2 + (t-u)^2 + (u-v)^2 + (v - 1)^2 - \frac15 = 0$
How can integer solutions exist if the last thing is a fraction?Are s, t, u, and v integers or real numbers
Isomorphism has done most of the hard work ...... Perhaps the following will keep the ball rolling for you:
$\displaystyle \therefore s^2 + (s-t)^2 + (t-u)^2 + (u-v)^2 + (v - 1)^2 = \frac{1}{5}$
So obviously (why?) each term on the left hand side has to be less than 1/5. This leads to a system of simultaneous inequalities ......
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