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!

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.
