Question

If the Pythagorean triples of one member is $8$, find the other two members.

• A. $15$ and $17$
• B. $14$ and $17$
• C. $15$ and $16$
• D. $11$ and $17$

Answer 1

The correct option is A $15$ and $17$

As we know $2m,$ $m^2+1,m^2-1$ form a Pythagorean triplet for any number $m>1$.
Let us assume $2m=8$
$\Rightarrow m=4$
Therefore, $m^2+1$ $=$ $4^2+1$
$\Rightarrow 16+1=17$
and $m^2-1$ $=$ $4^2-1$
$\Rightarrow 16-1=15$
Hence, the other two members are $15$ and $17$.