Question

Show that \[\left( {{m^2} - 1} \right),\left( {2m} \right),\left( {{m^2} + 1} \right)\] always form a Pythagorean triplet.

Answer 1

Hint: When the length of the side of a right triangle satisfies the Pythagoras theorem, these three numbers are known as Pythagorean triplets or triples. According to Pythagoras theorem, ${a^2} + {b^2} = {c^2}$.

Complete step-by-step answer:
Let $x = {m^2} - 1$ ,$y = 2m$ , and $z = {m^2} + 1$
To show x, y and z are Pythagorean triplet .So, we have to prove ${x^2} + {y^2} = {z^2}$ .
Now, $LHS = {x^2} + {y^2}$
Put the value of x and y in LHS
$ \Rightarrow {x^2} + {y^2} = {\left( {{m^2} - 1} \right)^2} + {\left( {2m} \right)^2}$
Use identity, ${\left( {p - q} \right)^2} = {\left( p \right)^2} - 2pq + {\left( q \right)^2}$
$
   \Rightarrow {x^2} + {y^2} = {\left( {{m^2}} \right)^2} - 2\left( {{m^2}} \right)\left( 1 \right) + {\left( 1 \right)^2} + 4{m^2} \\
   \Rightarrow {x^2} + {y^2} = {m^4} - 2{m^2} + 1 + 4{m^2} \\
   \Rightarrow {x^2} + {y^2} = {m^4} + 2{m^2} + 1 \\
 $
Use identity, ${\left( p \right)^2} + 2pq + {\left( q \right)^2} = {\left( {p + q} \right)^2}$
$
   \Rightarrow {x^2} + {y^2} = {\left( {{m^2}} \right)^2} + 2\left( {{m^2}} \right)\left( 1 \right) + {\left( 1 \right)^2} \\
   \Rightarrow {x^2} + {y^2} = {\left( {{m^2} + 1} \right)^2} \\
  LHS = {\left( {{m^2} + 1} \right)^2}............\left( 1 \right) \\
$
Now, $RHS = {z^2}$
Put value of z in RHS
$
   \Rightarrow {z^2} = {\left( {m{}^2 + 1} \right)^2} \\
  RHS = {\left( {m{}^2 + 1} \right)^2}.........\left( 2 \right) \\
 $
From (1) and (2) equation, LHS=RHS
Now, it's proven ${x^2} + {y^2} = {z^2}$ .
So, we can say \[\left( {{m^2} - 1} \right),\left( {2m} \right),\left( {{m^2} + 1} \right)\] always form a Pythagorean triplet.

Note: Whenever we face such types of problems we use some important points. First we assume triplets are sides of the right triangle and apply Pythagoras theorem then after using some algebraic identities if Pythagoras theorem satisfies. So, we can assume triplets are Pythagorean triplet.