<s>[INST] Find real $a$, $b$, $c$, $d$ satisfying $(1-a)^2+(a-b)^2+(b-c)^2+c^2=\frac{1}{4}$ 
If real numbers $a$, $b$, $c$, $d$ satisfy
  $$(1-a)^2+(a-b)^2+(b-c)^2+c^2=\frac{1}{4}$$ 
  then find $(a,b,c,d)$.

What I try: 
$$1+2a^2+2b^2+2c^2-2a-2ab-2bc=\frac{1}{4}$$
$$8a^2+8b^2+8c^2-8a-8ab-8bc+7=0$$
How do I solve it? Help me, please. 
 [/INST] Let $(p,q,r,s) = (1-a,a-b,b-c,c)$, we have $p+q+r+s = 1$.
We are given
$$p^2 + q^2 + r^2 + s^2 = \frac14$$
This leads to
$$\begin{align} &\;\left(p - \frac14\right)^2 +
\left(q - \frac14\right)^2 +
\left(r - \frac14\right)^2 +
\left(s - \frac14\right)^2\\
= &\; (p^2+q^2+r^2+s^2) - \frac12(p+q+r+s) + \frac14\\
= &\; \frac14 - \frac12 + \frac14\\
= &\; 0
\end{align}
$$
As a result, $$p = q = r = s = \frac14\quad\implies\quad (a,b,c) = \left(\frac34,\frac12,\frac14\right)$$
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