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Problem Of The Week #455 February 15th 2021

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anemone

MHB POTW Director
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Feb 14, 2012
3,894
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
Congratulations to the following members for their correct solution, which you can find below(Cool):

1. kaliprasad
2. Opalg

Solution from kaliprasad :
We are given
$xz-2yt=3\cdots(1)$
$xt+yz=1\cdots(2)$
square (1)and (2) to get
$x^2z^2 - 2xyzt + 4y^2t^2 = 9\cdots(3)$
$x^2t^2 -+ xyzt + y^2 x^2 = 2\cdots(4)$
from (3) and (4) we have
$x^2z^2 - 2xyzt + 4y^2t^2 + 2(x^2t^2 + xyzt + y^2 x^2) = 11$
or $x^2z^2 + 4y^2t^2 + 2 x^2y^2 + 2 y^2 x^2 = 11$
or $(x^2 + 2 t^2) (z^2 + 2y^2) = 11$
as LHS is product of 2 positive terms and RHS is 11 prime so we have 2 possibilities
$x^2 + 2 t^2 = 1$
$z^2 + 2y^2 = 11$
which gives the solution $x = \pm 1, t= 0, z = \pm 3, y = \pm 1$
Or
$x^2 + 2 t^2 = 11$
$z^2 + 2y^2 = 1$
which gives the solution $x = \pm 3, t= \pm 1, z = \pm 1, y = 0$
The above is based on the fact that $ 11 = 3^2 + 2 * 1^2$ and $1 = 1^2 + 2 * 0^2$
By checking the cases one by one we get the solution sets
$(x,y,z,t) = (1,0,3,1)$ or $(3,1,1,0)$ or $(-1,0,-3,-1)$ or $(-3,-1,-1,0)$

Solution from Opalg :
Remark: This solution uses university-level algebra so if you are a high school student, this might be beyond your knowledge to comprehend it.(Nod)

The Euclidean domain $\Bbb{Z}[\sqrt{-2}] = \{a+b\sqrt{-2}: a,b\in\Bbb{Z}\}$ has a multiplicative norm given by $N\bigl(a+b\sqrt{-2}\bigr) = a^2 + 2b^2$.

The given problem can be stated in the form $\bigl(x + y\sqrt{-2}\bigr) \bigl(z + t\sqrt{-2}\bigr) = 3 + \sqrt{-2}$. But $N\bigl(3+\sqrt{-2}\bigr) = 11$, which is a prime in $\Bbb{Z}$. It follows that the only factors of $3 + \sqrt{-2}$ are itself and $1$, and their negatives.Therefore there are four solutions to the problem, namely those in this table: $$\begin{array}{c|c|cccc}x + y\sqrt{-2} & z + t\sqrt{-2} &x&y&z&t \\ \hline 3 + \sqrt{-2} & 1 & 3&1&1&0 \\ 1 & 3 + \sqrt{-2} & 1&0&3&1 \\ -3 - \sqrt{-2} & -1 & -3&-1&-1&0 \\ -1 & -3 - \sqrt{-2} & -1&0&-3&-1 \end{array}.$$
 
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