$a$ , $b$ , $c$ , and $d$

${ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 }=ab+bc+cd+da$

${ a }^{ 2 }-ab+{ b }^{ 2 }+{ c }^{ 2 }-bc-cd+{ d }^{ 2 }-da=0\\ \frac { { a }^{ 2 } }{ 2 } -ab+\frac { { b }^{ 2 } }{ 2 } +\frac { { b }^{ 2 } }{ 2 } -bc+\frac { { c }^{ 2 } }{ 2 } +\frac { { c }^{ 2 } }{ 2 } -cd+\frac { { d }^{ 2 } }{ 2 } +\frac { { a }^{ 2 } }{ 2 } +\frac { { d }^{ 2 } }{ 2 } -da=0\\ \frac { 1 }{ 2 } \left\{ \left( { a }^{ 2 }-2ab+{ b }^{ 2 } \right) +\left( { b }^{ 2 }-2bc+{ c }^{ 2 } \right) +\left( { c }^{ 2 }-2cd+{ d }^{ 2 } \right) +\left( { d }^{ 2 }-2ad+{ a }^{ 2 } \right) \right\} =0\\ \frac { 1 }{ 2 } \left\{ { (a-b) }^{ 2 }+{ (b-c) }^{ 2 }+{ (c-d) }^{ 2 }+{ (d-a) }^{ 2 } \right\} =0\\ { (a-b) }^{ 2 }+{ (b-c) }^{ 2 }+{ (c-d) }^{ 2 }+{ (d-a) }^{ 2 }\quad =\quad 0$

So that

$a - b = 0 \Rightarrow a = b$

$b - c= 0 \Rightarrow b = c$

$c - d = 0 \Rightarrow c = d$

$d - a = 0 \Rightarrow d = a$

$\therefore a = b = c = d$

Then $a - b + c - d = 0$ .

We will use $AM-GM$ inequality to show that $a=b=c=d$ .

with $AM-GM$ , we have $a^2+b^2+c^2+d^2= \frac{a^2+b^2}{2} +\frac{b^2+c^2}{2} +\frac{c^2+d^2}{2} +\frac{d^2+a^2}{2} \\ \geq \sqrt{a^2b^2} +\sqrt{b^2c^2} +\sqrt{c^2d^2} +\sqrt{d^2a^2} = ab + bc + cd +ad$

with equality hold when $a=b=c=d$ which is same as the problem.

therefore, $\boxed{a-b+c-d = 0}$