Question

Find the value of the largest member of a Pythagorean triples if the value of the smallest member is $18$.

• A. $9$
• B. $80$
• C. $82$
• D. $81$

Answer 1

The correct option is C $82$

If the positive integers $a$, $b$ and $c$ form a Pythagorean triples, then the condition it needs to be satisfied is that $a^2+b^2=c^2.$ As examples, individually, $(3,4,5)$ and $(5,12,13)$ are Pythagorean triples because
$3^2+4^2=5^2(\because 9+16=25)$ and
$5^2+{12}^2={13}^2(\because 25+144=169).$

For any positive integer $n$, the numbers that form a Pythagorean triples would be:
$2n$, $n^2-1$ and $n^2+1$ because

$(2n{)}^2+(n^2-1{)}^2$
$=4n^2+(n^2{)}^2-2n^2+1$
$=n^4+2n^2+1$
$=(n^2{)}^2+2.n^2.1+1^2$
$=(n^2+1{)}^2$
$\Rightarrow (2n{)}^2+(n^2-1{)}^2=(n^2+1{)}^2$

Hence, the smallest member of the Pythagorean triples $(2n,n^2-1,n^2+1)$ would be $2n$.

According to the question, the smallest value of the Pythagorean triples is $18$.

$\therefore 2n=18$
$\Rightarrow \frac{2n}{2}=\frac{18}{2}$
$\Rightarrow n=9$

$\bullet$ Smallest value of the Pythagorean triples is $\underset{–}{18}$.

$\bullet$ Middle value of the Pythagorean triples:
$n^2-1$
$=9^2-1$
$=81-1$
$=\underset{–}{80}$

$\bullet$ Largest value of Pythagorean triples:
$n^2+1$
$=9^2+1$
$=81+1$
$=\underset{–}{82}$

Hence, the corresponding Pythagorean triples is $(18,80,82)$. And, the largest value of the Pythagorean triples would be $82$.

Therefore, option (c.) is the correct answer.