New Year 2017 Riddle (3 quadratic Diophanitine equations) What is the smallest prime number n which can be formulated with all of the
three following integer equations?
$$u^2 +  4^2 \cdot v^2 = n$$
$$w^2 +  9^2 \cdot x^2 = n$$
$$y^2 + 11^2 \cdot z^2 = n$$
Using the Theorem:


  If a prime can be expressed as sum of two squares then the representation is unique.
  Proof


Then we have that the three expressions are the same.

Then for minimize $n$ we need find a prime  that can be expressed as any of this forms:
$$4^2\cdot 11^2 r^2+9^2 j^2$$ or $$4^2\cdot 9^2 l^2+11^2 f^2$$
But $36^2+11^2=1417$ isn't a prime and by coincidences of the life $44^2+9^2=2017$ is a prime number. We are done.
$$ax_1^2+bx_2^2=ax_3^2+bx_4^2=ax_5^2+bx_6^2$$

The solution is easy to write.

$$x_1=t(bp^2+ak^2)(bs^2+ak^2)$$

$$x_2=y(bp^2+ak^2)(bs^2+ak^2)$$

$$x_3=((bp^2-ak^2)t-2bpky)(bs^2+ak^2)$$

$$x_4=((bp^2-ak^2)y+2apkt)(bs^2+ak^2)$$

$$x_5=((bs^2-ak^2)t-2bsky)(bp^2+ak^2)$$

$$x_6=((bs^2-ak^2)y+2askt)(bp^2+ak^2)$$

Here the representation of 3 options, but it is easy to see that can be written in the form of a combination with any number of options.