if right-handed-sign is \(a^2b^2c^2\), you can assume it.

Because the role of three numbers a,b,c is the same so suppose c≤b≤a

With [a=0b=0

 then a=b=c=0

(satisfy)

With a,b≠

0 we have:a2b2≤3a2⇒b2≤3⇒b∈{−1,1}

With b=-1 then a2+1+c2=a2⇒c2+1=0

(unsatisfactory)

With b=1 then a2+1+c2=a2⇒c2+1=0

(unsatisfactory)

Conclude:Only have a=b=c=0 satisfy the problem

+) With a = b = c = 0 (satisfy)

+) Consider a,b,c \(\ne\) 0 

     -) With a,b,c are three odd numbers

=> (a² + b² + c²) / 8 (balance 3) 

      (Because with x is an odd number , square of x divided for 8 will balance 1, we can prove it if we use formula of odd numbers)

By the way , at right hand side (a².b²) / 8 (balance 1) 

It's contradictory

So with a,b,c are three odd numbers (removed) 

     -) With a,b,c are three even numbers 

=> We decided \(\left\{{}\begin{matrix}a^2=2^n.x\\b^2=2^m.y\\c^2=2^p.z\end{matrix}\right.\)

=> \(a^2b^2\equiv0\) (mod 2^{m + n})  <=>   \(a^2b^2⋮2^{m+n}\)

But the left hand side is not 

So with a,b,c are three even numbers (removed)

    -) With a, b, c is not the same kind , we have this table 

Change the kind of three , it's not satisfy 

So a = b = c = 0 

Because the role of three numbers a,b,c is the same so suppose \(c\le b\le a\)

With \(\left[{}\begin{matrix}a=0\\b=0\end{matrix}\right.\) then \(a=b=c=0\)(satisfy)

With a,b\(\ne\)0 we have:\(a^2b^2\le3a^2\Rightarrow b^2\le3\)\(\Rightarrow b\in\left\{-1,1\right\}\)

With b=-1 then \(a^2+1+c^2=a^2\Rightarrow c^2+1=0\)(unsatisfactory)

With b=1 then \(a^2+1+c^2=a^2\Rightarrow c^2+1=0\)(unsatisfactory)

Conclude:Only have a=b=c=0 satisfy the problem