Candy Corn Concentric Circles

We note that the areas of the white, yellow and orange regions depend only on $z_1, (z_2+z_3), z_4$ in fact where the sign of equality indicates that the three regions are equivalent in the respective figures, therefore since the quaterne of the form $(a, a, a, a)$ $a\in N$ work (except for the condition on the gcd), the quaterne of the form $(a, a-k, a+k, a)$ $a\in N$ $k\in Z$ $|k|<a$ also work, now it is sufficient to choose k such that $gcd(z_1, z_2, z_3, z_4)=1$ i.e. $k=a-1$ .

Let $z_1=a,$ $z_1+z_2+z_3=b,$ and let $z_1+z_2+z_3+z_4=c.$ Then, we have:

$\begin{aligned} a^2+c^2-b^2 &= b^2-a^2 \\ c^2 &= 2b^2-2a^2 \\ c^2 &= 2(b-a)(b+a) \end{aligned}$

Given that $a,$ $b,$ and $c$ are integers, we note the following:

• $(b-a)(b+a) = 2n^2,$ where $n$ is a positive integer.
• Since $(b-a)(b+a)$ is even and $b$ and $a$ are integers, it follows that $(b-a)$ and $(b+a)$ must also be even.
• Then, it follows that $(b-a)(b+a)=8m^2,$ where $m$ is a positive integer.

Now a bit of guess-and-check with different values of $m$ to find a viable solution. With $8m^2=72,$ $(b-a) = 4,$ and $(b+a)=18,$ we have:

$\begin{aligned} a &= 7 \\ b &= 11 \\ c &= 12 \end{aligned}$

This gives $(z_1,z_2,z_3,z_4)=(7,1,3,1)$ or $(7,3,1,1).$ Thus, it is not necessary that $z_1=z_2=z_3=z_4.$

$(z_1, z_2, z_3, z_4) = (2,1,3,2)$ works, so, "No"

I am assuming that $z_2, z_3, z_4$ are distances between circumferences of circles, while $r_1$ is the radius of the smallest circle.

Suppose that (a, a, a, a) is a solution and this quanterne satisfies gcd = 1. Since the gcd = a, then a = 1. But by congruence, any multiple of (1, 1, 1, 1) is also a solution. So both hypotheses cannot be true.

A counterexample tells the option.--- NO

Z1=3, Z2=1 Z3=5 Z4=3

Here note the following, If

      Z1=Z4=k
  And Z3+Z2=2k

  all solution of   (Z1,Z2,Z3,Z4,k)  will satisfy the condition.

I just made a range of counterexamples, if z1=z4 and z1+z4=z2+z3 this will be no different from them all being equal. I, for example, used z1=1000, z2=1, z3=1999 and z4=1000, the only divisor of 1=z1 is 1 so the gcd must be 1 but it makes the same yellow, orange and white spaces as z1=z2=z3=z4=1000 which is just a scaled z1=z2=z3=z4=a for all a, so either this works or z1=z2=z3=z4 is wrong (and hence not true).

Start with the working solution given, and then increase $z_2$ and decrease $z_3$ such that $z_2 + z_3$ remains constant