# Understand the problem

Suppose that real numbers and satisfy the following equations:

Show that must be equal to or .

Show that must be equal to or .

*Note:*It is not required to show the existence of such numbers .##### Source of the problem

Germany MO 2019, Problem 6

##### Topic

Algebra, Simultaneous Equations

##### Difficulty Level

6/10

##### Suggested Book

Challenges and Thrills of Pre College Mathematics

# Start with hints

Do you really need a hint? Try it first!

Observe that x = y = z = 1 gives a valid solution of the set of equations. In this case s = x+y+z = 3. Now, observe one thing that this set of equations is symmetric in (x,y,z). Observe that we are required to comment on (x+y+z). Rewriting the equations as:

and then summing gives us that Our aim will be to reduce all the equations into a single variable. ( maybe a polynomial ). Let’s consider the case, where all of x,y,z is not 1.

and then summing gives us that Our aim will be to reduce all the equations into a single variable. ( maybe a polynomial ). Let’s consider the case, where all of x,y,z is not 1.

From now on we consider . This also gives Solving the first expression then plugging this into the second two gives:

as z is not equal to 1. Plugging the latter into the former and simplifying gives:

as z is not equal to 1. Plugging the latter into the former and simplifying gives:

Plugging the latter into the former and simplifying gives:

Now, observe that we already know z = 1 is a solution. This gives rise to

Now, observe that we already know z = 1 is a solution. This gives rise to

Observe that the polynomial we have got in terms of z is also satisfied by x,y,z as the equations are symmetric in x,y,z. Hence we can claim that \( t^3 – 7t^2 + 7t + 7 = 0 \) has three solutions x,y,z. Hence, \( t^3 – 7t^2 + 7t + 7 = (t-x)(t-y)(t-z)\). Therefore, by Vieta’s formula, x+y+z = 7. QED.

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