These are some of the Regional Spanish Mathematical Olympiad

1 We have a rectangle of $2 * n$ squares, in $n$ columns of 2 squares each. We have 3 colors to paint the squares, and we want that every two squares that share a side don't have the same color. How many diferent ways are there?

2 Find the real solutions of $\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx} = \frac{1}{6^{2}}$$\frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy} = \frac{7^{2}}{6^{2}}$$\frac{1}{(xy)^{2}}+\frac{1}{(yz)^{2}}+ \frac{1}{(zx)^{2}} = \frac{7^{2}}{36^{2}}$

3 For every positive integer $n \geq 1$ we denote $a_{n} = n^4 + n^2 + 1$. Find the greatest common divisor of $a_{n}$ and $a_{n+1}$ in function of $n$. Note by Jordi Bosch
5 years, 5 months ago

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I like question 2!

Staff - 5 years, 5 months ago pictureFor the square $ABCD$ we have three choices.For the square $BCFE$ we have two choices.Similarly,for the square $GDCH$ we have two choices.Now,for the square $HCIF$ we have two choices or one choice depending on the fact whether the squares,$GDCH$ and $BCFE$ have the same colour or different colours.WHAT TO DO NOW?

- 5 years, 5 months ago

The problem quickly follows if you consider that a row is already painted and you study the possibilities for next row.

- 5 years, 5 months ago

For 1. I got $2*3^n$,

for 2. I got $(x,y,z) = (-2,-3,6)$ and permutations and

for 3. I got $n^2 + n + 1$ if $7 \not | 2n+1$ and $7(n^2 + n + 1)$ if $7|2n+1$

- 5 years, 5 months ago

Under which condition does $7|2n+1$?

- 5 years, 5 months ago

When $2n + 1 \equiv 0 \pmod7$ or $2n \equiv 6 \pmod7$ or $n \equiv 3 \pmod7$

- 5 years, 5 months ago

Nicely done! that is if $n \equiv 3 \pmod{7}$ then the greatest common divisor of $a_{n}$ and $a_{n+1}$ is $7(n^2+n+1)$, else the gcd between $a_{n}$ and $a_{n+1}$ is $(n^2+n+1)$.

- 5 years, 5 months ago

- 5 years, 5 months ago

for 2 i got the solutions: (x.y.z)=( -2,-3,6)

- 5 years, 5 months ago

Correct :)

- 5 years, 5 months ago

Are there any other solutions?

Staff - 5 years, 5 months ago

We can make a cubic polynomial with x,y,z as root, we get can get 6 pairs of x,y,z(3 roots all real) by interchanging roots

$(-2,-3,6)$

$(-2, 6, -3)$

$(-3,-2,6)$

$(-3,6,-2)$

$(6,-2,-3)$

$(6,-3,-2)$

- 5 years, 5 months ago

no as @krishna sharma says we can obtain the values for x+y+z, xy+yz+xz and xyz and then declare a polynomial with roots x,y,z and place the coefficient values which will be x+y+z, xy+yz+xz and xyz and then use factor theorem to get the result

- 5 years, 5 months ago

I want learn these sums with enthusiasm.

- 5 years, 5 months ago