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If x, y and z are real numbers such that: x + y + z = 5 and xy + yz + zx = 3. What is the largest value that any of the numbers can have?

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Roots and Coefficients

If xyz = 1 and x+y+z =1/x + 1/y + 1/z show that at least one of these numbers must be 1. Now for the complexity! When are the other numbers real and when are they complex?

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Pair Squares

The sum of any two of the numbers 2, 34 and 47 is a perfect square. Choose three square numbers and find sets of three integers with this property. Generalise to four integers.

System Speak

Stage: 5 Challenge Level: Challenge Level:1

We received lots of solutions to this problem! Thank you to Lucas from Midhurst Rother College, Adithya, Amrit, Agathiyan and Guruvignesh from Hymers College, Moosa from Forest Gate Community School, Tim from Gosforth Academy, Bruno, Chris from Rastrick High School, Se Sun from Notre Dame High School, Aidan from Bonners Church of England Primary School, Pablo from King's College Alicante in Spain, Anson from Peak School in Hong Kong, Shah from MacKillop Catholic Regional College in Australia, Anton from Chairo Christian School in Australia and Chris from Reporoa College in New Zealand for sending us your solutions.

Moosa, Pablo and Tim used similar techniques to solve this problem. Here is Tim's method:

We can rearrange and substitute the equations to find a solution:

$$ab=1 \iff b = \frac{1}{a} \\
bc=2 \iff \frac{c}{a} \iff c=2a \\
cd=3 \iff 2ad=3 \iff d=\frac{3}{2a} \\
de = 4 \iff \frac{3e}{2a} = 4 \iff e=\frac{8a}{3} \\
ea = 6 \iff \frac{8a^2}{3}=6 \iff a^2=\frac{9}{4}$$Giving the solutions:
$$a=\frac{3}{2},\text{ or }-\frac{3}{2}$$If we subsitude these values back into the above equations, we can get the other values:
$$b=\frac{1}{a} \\
b=\frac{2}{3},\text{ or }-\frac{2}{3} \\
c=2a \\
c=3,\text{ or }-3 \\
d=\frac{3}{2a} \\
d=1,\text{ or }-1 \\
e=\frac{8a}{3} \\
e=4,\text{ or }-4$$So our two sets of values for ${a, b, c, d, a}$ are:
$$\left\{\frac{3}{2},\frac{2}{3},3,1,4\right\} \\
\text{and } \left\{-\frac{3}{2},-\frac{2}{3},-3,-1,-4\right\}$$

Se Sun used a different technique to solve this problem:

I decided to try multiplying each of the equations together. This produced:
$$\begin{align}
(ab)(bc)(cd)(de)(ea)&=1\times 2\times 3\times 4\times 6 \\
a^2b^2c^2d^2e^2&=144 \\
abcde&=\pm 12\qquad \text{Taking the square root of both sides}\\
cde&=\pm 12\qquad \text{Dividing by $ab=1$} \\
e&=\pm 4 \ \ \qquad \text{Dividing by $cd=3$}
\end{align}$$Now we can substitute into the other equations to find the other values:
$$de=4 \\
\pm 4d=4$$Therefore $d=\pm 1$.
$$cd=3 \\
\pm c=3$$Therefore $c=\pm 3$.
$$bc=2 \\
\pm 3b=2$$Therefore $b=\pm\frac{2}{3}$.
$$ab=1 \\
\pm\frac{2}{3}a=1$$Therefore $a=\pm\frac{3}{2}$.

Therefore $a=\frac{3}{2},b=\frac{2}{3},c=3,d=1,e=4$ or $a=-\frac{3}{2},b=-\frac{2}{3},c=-3,d=-1,e=-4$.