Solve system of eqns: xz-2yt=3, xt+yz=1; Find x2+y2+z2+t2 (x,y,z,t integers)

[Steven G]

x, y, z and t are all integers satisfying

xz-2yt = 3
xt + yz = 1

Find x²+y²+z²+t²

I tried squaring the equations but nothing seems to work.
It looks like it needs a unique trick to solve this.

Update: I used the fact that x and y are coprime (since xt + yz = 1) but still have not gotten any real results. I did notice by inspection that (x,y,z,t) = (3,1,1,0) works but that is kind of cheating. Besides I am sure that there are more solutions.

 

[daon2]

I failed to find a trick as well, and decided to try cases. Though I spent an annoying amount of time finding that


\(\displaystyle x^2+y^2 + z^2+t^2 = \dfrac{10z^2+6zt+13t^2 + (z^2+t^2)(z^2+2t^2)^2}{(z^2+2t^2)^2} \)

:roll:



Cases:


Since \(\displaystyle xt+yz=1, (x,y)=(x,z)=(t,y)=(t,z)=1\)


If We only look for non-negative integers, that means either:


...1) xt=1, yz=0... (since yz>0 contradicts the second equality)
........a) x=t=1 and y=0
........b) x=t=1 and z=0

...2) xt=0, yz=1.. (since xt>0 contradicts the second equality)
........a) x=0 and y=z=1
........b) t=0 and y=z=1


Cases 1a) and 2b) are the only that satisfy the first equation and consequently give z=3, and x=3 respectively.


Therefore the only nonnegative solutions are (1,0,3,1) and (3,1,1,0). There are also symmetric nonpositive solutions (-1,0,-3,-1), (-3,-1,-1,0). In all cases, the requested value is 11.


I doubt there are others but have not yet thought about how to prove over all integers that these are the only four.