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Find out the number of positive integral values of $a$ subjected to the equations $a+b+c=8$ and $ab+bc+ca=12$ where $a,b,c$ are real numbers.

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Hint: Expand the whole square of $a+b+c$ and substitute the given values from the question. The use trial and error method by putting different squares in the resultant equation.\[\]

Complete step-by-step answer:
The given equations are\[\]
$\begin{align}
  & a+b+c=8............\left( 1 \right) \\
 & ab+bc+ca=12....\left( 2 \right) \\
\end{align}$\[\]
First we expand the whole square of $a+b+c$ by multiplying with itself. That means \[\]
${{\left( a+b+c \right)}^{2}}=\left( a+b+c \right)\left( a+b+c \right)$ \[\]
Now we open the first bracket in order and multiply $a,b,c$ with $a+b+c$ separately and get\[\]
\[{{\left( a+b+c \right)}^{2}}=\left( a+b+c \right)\left( a+b+c \right)=a\left( a+b+c \right)+b\left( a+b+c \right)+c\left( a+b+c \right)\]
We open the second brackets in order and multiply $a,b,c$ individually with $\left( a+b+c \right)$ and get\[\]
$\begin{align}
  & {{\left( a+b+c \right)}^{2}}=a\cdot a+a\cdot b+a\cdot c+b\cdot a+b\cdot b+b\cdot c+b\cdot a+c\cdot a+c\cdot b+c\cdot c \\
 & ={{a}^{2}}+ab+ac+ba+{{b}^{2}}+bc+ba+ca+cb+{{c}^{2}} \\
\end{align}$ \[\]
We separate by collecting the square terms and product terms and get, \[\]
${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+ab+ba+bc+cb+ac+ca$ \[\]
We use the commutative property of multiplication where change in order of multiplication does not the product value.\[\]
$\begin{align}
  & {{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+ab+ba+bc+cb+ac+ca \\
 & ={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+ab+ab+bc+bc+ac+ca \\
 & ={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca \\
 & ={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\left( ab+bc+ca \right) \\
\end{align}$ \[\]
Substituting the values of $a+b+c$ and $ab+bc+ca$ in the above equation we get,
$\begin{align}
  & {{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\left( ab+bc+ca \right) \\
 & \Rightarrow {{\left( 8 \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\left( 12 \right) \\
\end{align}$ \[\]
On simplification we get,\[\]
$\begin{align}
  & 64={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+24 \\
 & \Rightarrow 64-24={{a}^{2}}+{{b}^{2}}+{{c}^{2}} \\
 & \Rightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}=40.....(3) \\
\end{align}$\[\]
If the square root of a number is an integer then the number is called the perfect square.
We have found that ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}$ to be 40. If $a$is an integer then ${{a}^{2}}$ is also a positive integer and a perfect square . Similarly ${{b}^{2}}$ and ${{c}^{2}}$ are integers and perfect squares.\[\]
  Observe that 40 is now the sum of three perfect squares. So the possible values of ${{a}^{2}},{{b}^{2}},{{c}^{2}}$ are 0, 1, 4, 16, 25, 36.
Now the only way 40 can be expressed as a sum is three perfect squares when the perfect squares are taken from the set $\left\{ 0,4,36 \right\}$.
We first take ${{b}^{2}}=0$ and ${{c}^{2}}=4$ then from equation (3)
\[{{a}^{2}}+0+4=36\Rightarrow a=6\text{ or }-6\]
If we take ${{b}^{2}}=4$ and ${{c}^{2}}=0$ we are going to obtain the same result.
Now we take ${{b}^{2}}=36$ and ${{c}^{2}}=0$ and put in the equation(3)
\[{{a}^{2}}+0+36=4\Rightarrow a=2\text{ or }-2\]
If we take ${{b}^{2}}=0$ and ${{c}^{2}}=36$ we are going to obtain the same result.
Now we take ${{b}^{2}}=36$ and ${{c}^{2}}=1$ put in the equation(3)
\[{{a}^{2}}+4+36=0\Rightarrow a=0\]
If we take ${{b}^{2}}=4$ and ${{c}^{2}}=4$ we are going to obtain the same result.
\[\]The possible values of $a$ are 6, -6, 2, -2, 0. We reject the non-negative integral values as asked in the question. \[\] The values that remain are 6,2.\[\]
So the number of positive integral values of $a$ is 2.

Note: The questions test the concept of perfect squares. We need to be careful while placing values in equations. It is also to be noted that the square of the same positive and negative integers is the same.