For nonnegative real numbers $x,y,z$ and $t$ we know that $|x-y|+|y-z|+|z-t|+|t-x|=4$.
Find the minimum of $x^2+y^2+z^2+t^2$.

[i]proposed by Mohammadmahdi Yazdi, Mohammad Ahmadi[/i]
1. Given the equation \( |x-y| + |y-z| + |z-t| + |t-x| = 4 \), we need to find the minimum value of \( x^2 + y^2 + z^2 + t^2 \) for nonnegative real numbers \( x, y, z, t \).

2. We start by applying the Cauchy-Schwarz inequality in the form:
   \[
   (a_1^2 + a_2^2 + a_3^2 + a_4^2)(b_1^2 + b_2^2 + b_3^2 + b_4^2) \geq (a_1b_1 + a_2b_2 + a_3b_3 + a_4b_4)^2
   \]
   where \( a_i = 1 \) and \( b_i \) are the absolute differences \( |x-y|, |y-z|, |z-t|, |t-x| \).

3. Applying this to our problem, we have:
   \[
   (x^2 + y^2 + z^2 + t^2)(1^2 + 1^2 + 1^2 + 1^2) \geq (|x-y| + |y-z| + |z-t| + |t-x|)^2
   \]
   Simplifying, we get:
   \[
   4(x^2 + y^2 + z^2 + t^2) \geq 4^2
   \]
   \[
   4(x^2 + y^2 + z^2 + t^2) \geq 16
   \]
   \[
   x^2 + y^2 + z^2 + t^2 \geq 4
   \]

4. To check if this bound is achievable, we need to find values of \( x, y, z, t \) such that \( |x-y| + |y-z| + |z-t| + |t-x| = 4 \) and \( x^2 + y^2 + z^2 + t^2 = 4 \).

5. Consider the values \( x = 1, y = 0, z = 1, t = 0 \):
   \[
   |x-y| = |1-0| = 1
   \]
   \[
   |y-z| = |0-1| = 1
   \]
   \[
   |z-t| = |1-0| = 1
   \]
   \[
   |t-x| = |0-1| = 1
   \]
   Adding these, we get:
   \[
   |x-y| + |y-z| + |z-t| + |t-x| = 1 + 1 + 1 + 1 = 4
   \]
   Also, calculating \( x^2 + y^2 + z^2 + t^2 \):
   \[
   x^2 + y^2 + z^2 + t^2 = 1^2 + 0^2 + 1^2 + 0^2 = 1 + 0 + 1 + 0 = 2
   \]

6. Therefore, the minimum value of \( x^2 + y^2 + z^2 + t^2 \) is indeed \( 2 \).

\[
\boxed{2}
\]