Solve for $a,b,c,d \in \Bbb R$, given that $a^2+b^2+c^2+d^2-ab-bc-cd-d+\frac 25 =0$ Today, I came across an equation in practice mock-test of my coaching institute, aiming for engineering entrance examination (The course for the test wasn't topic-specific, it was a test of complete high school mathematics). It was having four variables and only one equation. While analyzing my test paper, this is the only problem I (and my friends too) couldn't figure out even after giving this problem several hours. So I came here for some help.

Question : Solve for $a,b,c,d \in \Bbb R$, given that $$a^2+b^2+c^2+d^2-ab-bc-cd-d+\frac 25 =0$$

Since only one equation is given, there must be involvement of making of perfect squares, such that they all add up to $0$. Thus, resulting in few more equations. But how to?
I tried a lot of things, such as making $(a-b)^2 $ by adding the missing terms and subtracting again, but got no success.
Thanks!
 A: Let $F(a,b,c,d) = a^2+b^2+c^2+d^2-ab-bc-cd-d+\frac25$. 
With help of a CAS, one can verify
$$\begin{align}
F\left(\frac15+p,\frac25+q,\frac35+r,\frac45+s\right)
&= p^2 - pq + q^2 - qr + r^2 -rs + s^2\\
&= \frac12\left(p^2 + (p-q)^2 + (q-r)^2 + (r-s)^2 + s^2\right)\end{align}$$
If one set $(a,b,c,d)$ to $\left(\frac15+p,\frac25+q,\frac35+r,\frac45+s\right)$, one find
$$\begin{align}
F(a,b,c,d) = 0 
&\iff p = p-q = q-r = r-s = s = 0\\
&\iff p = q = r = s = 0
\end{align}
$$
This implies the equation at hand has a unique solution:
$$(a,b,c,d) = \left(\frac15,\frac25,\frac35,\frac45\right)$$
Update
About the question how I come up with this. I first write $F(a,b,c,d)$ as
$$\begin{align}
F(a,b,c,d) &= a^2 + b^2 + c^2 + d^2 - ab - bc - cd - da + d(a-1) + \frac25\\
&=  \frac12((a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2) + d(a-1) + \frac25
\end{align}\tag{*1}
$$
To simplify the term $d(a-1)$, I introduce $\lambda, \mu$ such that
$$\begin{cases}
d &= \frac12 + \lambda + \mu\\
a &= \frac12 + \lambda - \mu
\end{cases}
\quad\implies\quad d(a-1) = \lambda^2 - \left(\frac12+\mu\right)^2
$$
Now $d-a = 2\mu$ and $(a-b)^2 + (b-c)^2 + (c-d)^2 
\ge 3\left(\frac{d-a}{3}\right)^2
= \frac43 \mu^2$.
If one substitute this back into $(*1)$, one find
$$F(a,b,c,d) \ge \frac83\mu^2 + \lambda^2 - (\frac12 + \mu)^2 + \frac25
= \lambda^2 + \frac53\left(\mu - \frac{3}{10}\right)^2$$
In order for $F(a,b,c,d) = 0$, we need 
$$\lambda = 0,\quad\mu = \frac{3}{10}
\quad\text{ and }\quad(a-b)^2 + (b-c)^2 + (c-d)^2 = \frac13(d-a)^2$$
The last condition forces $d-c = c-b = b-a = \frac13(d-a)$ and leads to the solution $(a,b,c,d) = \left(\frac15,\frac25,\frac35,\frac45\right)$.
This is a little bit sloppy to describe, so I look at expansion
of $F(a,b,c,d)$ near the solution and obtain a simpler description of $F$ in terms of $p,q,r,s$.
A: Multiply by $2$ and rearrange to \begin{align*}(a-b)^2 + (b-c)^2 + (c-d)^2 + (d-a)^2 + 2ad - 2d + \frac{4}{5} = 0. \tag{$\star$}\end{align*} For fixed $a$ and $d$, the minimum value of $(a-b)^2 + (b-c)^2 + (c-d)^2$ is $\frac{(d - a)^2}{3}$,  with equality if and only if $a, b, c, d$ is an arithmetic progression by the lemma below, so the LHS of $(\star)$ is at least $$\frac{4}{3} (d-a)^2 + 2ad - 2d + \frac{4}{5} = \frac{4}{3} \left( a - \frac{d}{4} \right)^2 + \frac{5}{4} \left( d - \frac{4}{5}\right)^2 \tag{$\dagger$}.$$
But $(\dagger)$ is clearly non-negative, and it is zero if and only if $d = 4/5$ and $a = d/4 = 1/5$, but the LHS of $(\star)$ must be zero. From this, $b = 2/5$ and $c = 3/5$ follow, and there can be no other solution.
Lemma. For fixed $x_0$ and $x_n$, the sum $\sum_{i=1}^n (x_i - x_{i-1})^2$ is minimized when the $x_i$ form an arithmetic progression $x_i = \frac{n-i}{n} x_0 + \frac{i}{n} x_n$.
Proof. For $n = 2$, $(x_0 - x_1)^2 + (x_1 - x_2)^2$ can be rearranged as $$ \left( x_1 - \frac{x_0 + x_2}{2} \right)^2 +x_0^2 + x_2^2 - \frac{(x_0 + x_2)^2}{4}. $$
For $n > 2$, if some $x_k$ is not the midpoint of $x_{k-1}$ and $x_{k+1}$, then $(x_k - x_{k-1})^2 + (x_{k+1} - x_k)^2$ can be reduced by moving $x_k$ to the midpoint, leaving the other terms of $\sum_{i=1}^n (x_i - x_{i-1})^2$ alone. So if a minimum exists, it must have evenly spaced $x_i$. And proving that a minimum exists is simple: the possible values of $x_1, \ldots, x_{n-1}$ that can minimize $f(x_1, \ldots, x_{n-1}) = \sum_{i=1}^n (x_i - x_{i-1})^2$ can be bounded in some closed interval $[-R, R]$, and the image of a connected compact set $[-R, R]^n$ under a continuous function $f$ must be compact and connected (that is, a closed bounded interval).
This lemma can be interpreted physically as stating that the potential energy of a chain of $n$ identical springs with unstretched length zero, with the endpoints of the whole chain anchored, is minimized (and thus the forces at each spring endpoint are in equilibrium) when each spring is stretched equally. Here, $x_0$ and $x_n$ are the fixed endpoints, and $x_{i-1}$ and $x_i$ are the endpoints of the $i$th spring.
A: Your idea of turning the right side into a sum of perfect squares is a good one.  Observing that
$$
(a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2=2(a^2+b^2+c^2+d^2-ab-bc-cd-da)
$$
we multiply
$$
a^2+b^2+c^2+d^2-ab-bc-cd-d+\frac{2}{5}=0
$$
by $2$ and rewrite the result as
$$
(a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2+2ad-2d+\frac{4}{5}=0.
$$
We define $x=a-b$, $y=b-c$, and $z=c-d$, which gives $a=x+y+z+d$.  Substituting gives
$$
x^2+y^2+z^2+(x+y+z)^2+2d(x+y+z+d-1)+\frac{4}{5}=0,
$$
which can be rewritten as
$$
x^2+y^2+z^2+(x+y+z+d)^2+d^2-2d+\frac{4}{5}=0
$$
or
$$
x^2+y^2+z^2+(x+y+z+d)^2+(d-1)^2=\frac{1}{5}.
$$
So a sum of five perfect squares equals $\frac{1}{5}$.  We might hope for a solution in which each of the perfect squares is $\left(\pm\frac{1}{5}\right)^2$, and indeed we find that $x=y=z=-\frac{1}{5}$, $d=\frac{4}{5}$ provides such a solution.
Now we ask whether the solution can be perturbed.  Letting $x=-\frac{1}{5}+e$, $y=-\frac{1}{5}+f$, $z=-\frac{1}{5}+g$, $d=\frac{4}{5}+h$ and substituting, we get
$$
(-1/5+e)^2+(-1/5+f)^2+(-1/5+g)^2+(1/5+e+f+g+h)^2+(-1/5+h)^2=\frac{1}{5},
$$
which simplifies to
$$
e^2+f^2+g^2+h^2+(e+f+g+h)^2=0.
$$
This forces $e=f=g=h=0$, and therefore the solution is unique.
A: It's 
$$a^2-ab+\frac{b^2}{4}+\frac{3}{4}b^2-bc+\frac{1}{3}c^2+\frac{2}{3}c^2-cd+\frac{3}{8}d^2+\frac{5}{8}d^2-d+\frac{2}{5}=0$$ or
$$\left(a-\frac{b}{2}\right)^2+\frac{3}{4}\left(b-\frac{2c}{3}\right)^2+\frac{2}{3}\left(c-\frac{3d}{4}\right)^2+\frac{5}{8}d^2-d+\frac{2}{5}=0$$ or
$$\left(a-\frac{b}{2}\right)^2+\frac{3}{4}\left(b-\frac{2c}{3}\right)^2+\frac{2}{3}\left(c-\frac{3d}{4}\right)^2+\frac{5}{8}\left(d-\frac{4}{5}\right)^2=0,$$ 
which gives the answer. 
A: The expression is a quadratic form. Experimenting a bit you can put $v^{T}=(a, b, c, d, 1)$ and let A be the 5 by 5 matrix with 1, 1, 1, 1, $\frac{2}{5}$ down the lead diagonal, with -1, -1, -1, -1 above that and 0 everywhere else. Then the expression is $v^{T}Av = 0$. The expression is also $v^{T}A^{T}v = 0$ and better still it is $v^{T}(A+A^{T})v = 0$ where $A+A^{T}$ is a symmetric matrix.
In this form $(A+A^{T})v = 0$ has the unique solution v already given. It would be nice if it were obvious that this is the unique solution of $v^{T}(A+A^{T})v = 0$ but I can't see it. 
A: I think that the following is most natural way to solve this equation. $$a^2+b^2+c^2+d^2-ab-bc-cd-d+\frac 25 =0$$
Since it is quadratic on $a$ we have nonegative discriminant:
$$ b^2- 4(b^2+c^2+d^2-bc-cd-d+\frac 25) \geq 0$$
so 
$$4c^2-4c(b+d)+4d^2+3b^2-4d +\frac85 \leq 0$$
so 
$$4c^2-4c(b+d)+(b+d)^2-(b+d)^2+4d^2+3b^2-4d +\frac85 \leq 0$$
so 
$$(2c-b-d)^2+3d^2+2b^2-2bd-4d+\frac85 \leq 0$$
so
$$3d^2+2b^2-2bd-4d+\frac85 \leq 0\;\;\;/\cdot 2$$
so 
$$4b^2-4bd+d^2+5d^2-8d+\frac{16}5 \leq 0 \;\;\;/\cdot 5$$
so 
$$5(2b-d)^2+25d^2-40d+16 \leq 0 $$
so 
$$5(2b-d)^2+(5d-4)^2 \leq 0 $$
which means that $d=4/5$ and $b=2/5$ and $c=(b+d)/2=...$ and $a=...$
