Question

The Pythagorean triplets whose smallest number is 8

Answer 1

Hint: A Pythagorean triplet is a set of positive integers a, b and c that fit the rule ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$
 In a right-angle triangle, if P is perpendicular, B be base and H be the hypotenuse then we have a
${{H}^{2}}={{P}^{2}}+{{B}^{2}}$
H, P, B be the Pythagorean triplets. So, in order to find out Pythagorean triplet if on number is given use identity ${{(a+b)}^{2}}={{(a-b)}^{2}}+{{\left( 2\sqrt{ab} \right)}^{2}}$and compare it with the side of right-angled triangle.

Complete step-by-step answer:
As we know from Pythagoras theorem in a right-angled triangle.









${{H}^{2}}={{P}^{2}}+{{B}^{2}}$
Now as we know from identity
${{(a+b)}^{2}}={{(a-b)}^{2}}+4ab$
So, it can be written as
${{(a+b)}^{2}}={{(a-b)}^{2}}+{{\left( 2\sqrt{ab} \right)}^{2}}$
So, on comparing we can write the above as
$\begin{align}
  & a+b=H \\
 & a-b=P \\
 & 2\sqrt{ab}=B \\
\end{align}$
As hypotenuse is the largest side of the right-angled triangle. Now let us suppose that
 \[\begin{align}
  & \sqrt{a}=m \\
 & \sqrt{b}=n \\
\end{align}\]
Here we assume that $m>n$ , where m and n are integers, so we can write the above as
$\begin{align}
  & \Rightarrow {{m}^{2}}+{{n}^{2}}=H \\
 & {{m}^{2}}-{{n}^{2}}=P \\
 & 2mn=B \\
\end{align}$
Now suppose that the given number 8 is for base that is B which is least among triplets so we can write it as
$\begin{align}
  & 2mn=8\text{ } \\
 & \text{mn=4} \\
\end{align}$
Hence value of m and n can be
$m=1\text{ , n=4}$
or
$m=4,n=1$
or
$m=2,n=2$
But we assume $m>n$ and m and n are integers
So
$\begin{align}
  & m=4 \\
 & n=1 \\
\end{align}$
will satisfy the given condition
Hence, we can find the following values for Pythagorean triplets
$\begin{align}
  & H={{4}^{2}}+1=17 \\
 & P={{4}^{2}}-1=15 \\
 & B=8 \\
\end{align}$
So, triplet will be$(17,\text{ }15,\text{ 8)}$

Note: we can use direct formula to find out triplet as if triplet N is even triplet are
\[\begin{align}
  & {{\left( \dfrac{N}{2} \right)}^{2}}-1,\text{ }{{\left( \dfrac{N}{2} \right)}^{2}}+1\text{ }\!\!\And\!\!\text{ N} \\
 & \text{If N is odd the triplets are} \\
 & \left( {{\dfrac{N}{2}}^{2}}-0.5 \right)and\text{ }\left( {{\dfrac{N}{2}}^{2}}+0.5 \right)\text{ }\!\!\And\!\!\text{ N} \\
\end{align}\]
If a, b, and c are Pythagorean triplets then ka, kb and kc will be also a set of Pythagorean triples where k is a positive integer.