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Hint: First of all, take the two positive integers as variables. Then find their sum of squares of the integers and the product of the integers. Add up them and equate it to a perfect square to get the general equation.

Complete step-by-step answer:

Let \[x,y\] be two positive integers

The sum of the squares of the positive integers \[ = {x^2} + {y^2}\]

The product of the positive integers \[ = xy\]

The result obtained by adding the two positive integers product to the sum of their squares \[ = {x^2} + {y^2} + xy\]

Let the obtained result is equal to a perfect square of the form \[{z^2}\]

So, we have \[{x^2} + xy + {y^2} = {z^2}\]

Which can be written as

\[

\Rightarrow x\left( {x + y} \right) + {y^2} = {z^2} \\

\Rightarrow x\left( {x + y} \right) = {z^2} - {y^2} \\

\Rightarrow x\left( {x + y} \right) = \left( {z + y} \right)\left( {z - y} \right) \\

\]

Since \[x,y\] are positive integers, \[x,x + y\] are also positive hence their product i.e., \[x\left( {x + y} \right)\] is also positive.

Now put \[mx = n\left( {z + y} \right){\text{ and }}n\left( {x + y} \right) = m\left( {z - y} \right)\] where \[m,n\] are positive integers then we get

\[ \Rightarrow \dfrac{x}{{2mn + {n^2}}} = \dfrac{y}{{{m^2} - {n^2}}} = \dfrac{z}{{{m^2} + mn + {n^2}}}\] which is the required solution.

Thus, the general formula to express two positive integers which are such that the result obtained by adding their product to the sum of their squares is a perfect square is given by \[\dfrac{x}{{2mn + {n^2}}} = \dfrac{y}{{{m^2} - {n^2}}} = \dfrac{z}{{{m^2} + mn + {n^2}}}\].

Note: Here perfect square is a number, from a given number system, that can be expressed as the square of a number from the same number system. The obtained general formula is valid for all positive integers.

Complete step-by-step answer:

Let \[x,y\] be two positive integers

The sum of the squares of the positive integers \[ = {x^2} + {y^2}\]

The product of the positive integers \[ = xy\]

The result obtained by adding the two positive integers product to the sum of their squares \[ = {x^2} + {y^2} + xy\]

Let the obtained result is equal to a perfect square of the form \[{z^2}\]

So, we have \[{x^2} + xy + {y^2} = {z^2}\]

Which can be written as

\[

\Rightarrow x\left( {x + y} \right) + {y^2} = {z^2} \\

\Rightarrow x\left( {x + y} \right) = {z^2} - {y^2} \\

\Rightarrow x\left( {x + y} \right) = \left( {z + y} \right)\left( {z - y} \right) \\

\]

Since \[x,y\] are positive integers, \[x,x + y\] are also positive hence their product i.e., \[x\left( {x + y} \right)\] is also positive.

Now put \[mx = n\left( {z + y} \right){\text{ and }}n\left( {x + y} \right) = m\left( {z - y} \right)\] where \[m,n\] are positive integers then we get

\[ \Rightarrow \dfrac{x}{{2mn + {n^2}}} = \dfrac{y}{{{m^2} - {n^2}}} = \dfrac{z}{{{m^2} + mn + {n^2}}}\] which is the required solution.

Thus, the general formula to express two positive integers which are such that the result obtained by adding their product to the sum of their squares is a perfect square is given by \[\dfrac{x}{{2mn + {n^2}}} = \dfrac{y}{{{m^2} - {n^2}}} = \dfrac{z}{{{m^2} + mn + {n^2}}}\].

Note: Here perfect square is a number, from a given number system, that can be expressed as the square of a number from the same number system. The obtained general formula is valid for all positive integers.