Prove that $a^2  + b^2 + c^2 $ is not a prime number I am having difficulty solving this problem:
Let $a, b, c \in\mathbb{Z}$, $abc \neq 0$ and $a\neq c$ be such that
$$\frac{a}{c} = \frac{a^2+b^2}{c^2+b^2}.$$
Prove that $a^2 + b^2 + c^2$ is not a prime number.
Thanks in advance!
 A: $\rm\  a^2+b^2+c^2 = (a+c)^2-b^2 + 2\: (b^2-ac)\ $ so $\rm\:b^2\! = ac\:\Rightarrow\:$ it factors (difference of squares)
A: NOTE: The "solution" below addressed the original quesion, which was "Prove that $a^2 +b^2 +ab$ is not a prime number", and was later changed to its present form. In fact, it is the case that every prime congruent to $1$ (mod $3$) has the form $a^2 +b^2 +ab$ for integers $a$ and $b,$ while no prime congruent to $2$ (mod $3$) has this form. The former (well-kown) statement can be proved in a fashion rather similar to Euler's proof that every prime congruent to $1$ (mod $4$) is a sum of two integer squares. In this case, however, one works with the ring of Eisenstein integers, $R = \mathbb{Z}[\omega],$ where $\omega$ is a primitive (complex) cube root of unity. This is a principal ideal domain. If $p \equiv 1$ (mod $3$) is a rational prime, then the multiplicative group of the field $\mathbb{Z}/p\mathbb{Z}$ contains an element of order $3$. Hence there is an integer $n$ such that $p$ divides $n^{3}- 1,$ but $p$ does not divide $n-1.$ Then $p$ divides $n^{2}+n+1,$ which factors as $(n- \omega)(n-\omega^{2})$ in $R.$
Since $p$ does not divide either of the two factors in $R,$ we must conclude that $p$ is not a prime in $R.$ Hence there are integers $a,b,c,d$ such that $p = (a - b \omega)(c- d\omega)$ in $R,$ where neither $a-b\omega$ nor $c-d\omega$ are units in $R.$ Then multiplying this expression by its complex coinjugate , we see that $p^2 =  (a^2 +ab + b^2)(c^2 +cd +d^2).$ Now $a^2 +ab +b^2 \neq 1$ and $c^2 +cd +d^2 \neq 1$ as $a-b\omega$ and $c-d\omega$ are non-units in $R.$ Hence $a^2 +ab +b^2 = p$ (note that it is a positive quanitity). It is an easy exercise that if $q \equiv 2$ (mod $3$), then $q$ remains prime in $R,$ so $q$ can certainly not be written in the form $a^2 +ab+b^2$ for ratonal integers $a$ and $b.$
A: W.l.o.g we may assume that $a,c>0$. The equation
$$
\frac{a}{c}=\frac{a^2+b^2}{c^2+b^2}
$$
together with the assumption $a\neq c$ quickly gives us $ac=b^2$ as a corollary. Therefore we have
$$
a^2+b^2+c^2=a^2+ac+c^2
$$
and the extra condition that $ac=b^2$ must be a perfect square.
There are two main cases. If $gcd(a,c)>1$, then that common divisor is  also a divisor of $a^2+ac+c^2$, so this latter number won't be a prime. If the numbers $a$ and $c$ are coprime, then the equation $ac=b^2$ and unique factorization force both $a$ and $c$ to be squares. So we can assume that $a=p^2, c=q^2$ for some integers $p,q$. But then we see that
$$
a^2+b^2+c^2=p^4+p^2q^2+q^4=(p^2+q^2)^2-p^2q^2=(p^2+pq+q^2)(p^2-pq+q^2).
$$
Here $p\neq q$, so both these factors are $>1$, and the claim follows in this case, too.
A: if we arrange it  by another way like this  $(a*c^2+a*b^2)$=$(c*a^2+c*b^2)$ and then concatate  similar terms we get  $(a*c*(c-a))$=$b^2$*$(c-a)$  or $b^2$=$a*c$   so  $a^2$+$b^2$+$c^2$=$a^2$+$c^2$+$a*c$ which  already  is know that it is  prime  for some  variables
