Question

If $ x,y,z $ are positive integers and $ x+y+z=15,xy+yz+xz=72 $ then $ 3\le x\le 7 $ . \[\]
A. True\[\]
B. False \[\]

Answer 1

Hint: We use the algebraic identity $ {{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\left( ab+bc+ca \right) $ and find that $ {{x}^{2}}+{{y}^{2}}+{{z}^{2}}=81 $ . Since we are given $ x,y,z $ are positive integers we restrict our choice for $ {{x}^{2}},{{y}^{2}},{{z}^{2}} $ to perfect squares under 81. We use trial and error method to guess different values of $ {{x}^{2}},{{y}^{2}},{{z}^{2}} $ and check if they satisfy given conditions or not. \[\]

Complete step by step answer:
We know that the perfect square under 100 are 1, 4,9,16,25,36,49,64,81 because they are the squares of 1,2,3,4,5,6,7,8,9 respectively.\[\]
We know from algebraic identity of whole square of sum of three numbers $ a,b,c $ is given by
\[{{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\left( ab+bc+ca \right)\]
We are give in the question 3 unknowns and 2 equations
\[\begin{align}
  & x+y+z=15.......\left( 1 \right) \\
 & xy+yz+xz=72.........\left( 2 \right) \\
\end{align}\]
We use whole square of sum of three numbers identify by taking $ a=x,b=y,c=z $ to have;
\[{{\left( x+y+z \right)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2\left( xy+yz+zx \right)\]
We put the values as from equation (1) and (2) in the above step to have;
\[\begin{align}
  & {{\left( 15 \right)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2\left( 72 \right) \\
 & \Rightarrow 225={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+144 \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}+{{z}^{2}}=81 \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{9}^{2}} \\
\end{align}\]

We see in the left hand side of the equation we have three squares of positive integers and in the right hand side of the equation we have the square of 9. So all the three squares $ {{x}^{2}},{{y}^{2}},{{z}^{2}} $ will be the square of 9 and hence the unknowns $ x,y,z $ will also be less than 9. \[\]
So we need to use guess and error methods to find $ x,y,z $ . We guess different numbers from 1,2,3,4,5,6,7,8 for $ x,y,z $ and see if their squares $ {{x}^{2}},{{y}^{2}},{{z}^{2}} $ from 1, 4,9,16,25,36,49,64 will add up to 81. \[\]
We can express their sum of square as $ {{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{x}^{2}}+\left( {{y}^{2}}+{{z}^{2}} \right) $ . Since 81 is an odd number it can be expressed as the sum of an even and an odd number. We can take $ {{x}^{2}}=1 $ the choice for an odd number and try to express $ 81-1=80 $ as the sum of two squares. We can do that by choosing $ {{y}^{2}}=16,{{z}^{2}}=64 $ to get
\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}=1+16+64=81\]
We can also interchange the values for $ {{x}^{2}},{{y}^{2}},{{z}^{2}} $ from 1, 16, 64 but then we shall have $ x=1,y=4,z=8 $ which shall not satisfy the given condition $ x+y+z=15 $ because
\[x+y+z=1+4+8=13\ne 15\]
Now we choose $ {{x}^{2}}=9 $ and try to express $ 81-9=72 $ in terms of two squares . We can do that by only choosing $ {{y}^{2}}={{z}^{2}}=36 $ which will give us
\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}=9+36+36=81\]
So we have $ \left( x,y,z \right)=\left( 3,6,6 \right) $ .We also have the given condition satisfied
\[\begin{align}
  & x+y+z=3+6+6=15 \\
 & xy+yz+zx=3\cdot 6+6\cdot 6+6\cdot 3=72 \\
\end{align}\]
We can interchange also interchange the values for $ {{x}^{2}},{{y}^{2}},{{z}^{2}} $ from 9, 36, and 36 to have solution triples
\[\left( x,y,z \right)=\left( 3,6,6 \right),\left( 6,3,6 \right),\left( 6,6,3 \right).....\left( 3 \right)\]
We then choose $ {{x}^{2}}=25 $ but we do not any $ y,z $ such that $ {{x}^{2}}+{{y}^{2}}+{{z}^{2}}=81 $ because we are not get any $ y,z $ such that $ {{y}^{2}}+{{z}^{2}}=81-25=56 $ .
We then choose $ {{x}^{2}}=49 $ then we can choose $ {{y}^{2}}=16,{{z}^{2}}=16 $ to have $ \left( x,y,z \right)=\left( 7,4,4 \right) $ . We also have
\[\begin{align}
  & {{x}^{2}}+{{y}^{2}}+{{z}^{2}}=49+16+16=81 \\
 & x+y+z=7+4+4=15 \\
 & xy+yz+zx=7\cdot 4+4\cdot 4+4\cdot 7=72 \\
\end{align}\]
So $ \left( x,y,z \right)=\left( 7,4,4 \right) $ is the solution triple. We can interchange $ x,y,z $ to have possible solution triples
\[\left( x,y,z \right)=\left( 7,4,4 \right),\left( 4,4,7 \right),\left( 4,7,4 \right)......\left( 4 \right)\]
We see from results (3) and (4) that the all values of $ x $ lie in between 3 and 7. So it is true that $ 3\le x\le 7 $ .\[\]

Note:
 We know that a perfect square is a positive integer whose square root is also a positive integer. A square is always non-negative, since we are given $ x,y,z $ are positive integers out choices are restricted to perfect squares. We have used the inequality of square root $ a\le b\Rightarrow \sqrt{a}\le \sqrt{b} $ when we deduced that $ x,y,z $ will be less than $ \sqrt{81}=9 $ .