Show the following identities concerning Pythagorean triples:
(a) $(3n)^2 + (4n)^2 = (5n)^2$
(b) $(2n)^2 + (n^2− 1)^2 = (n^2 + 1)^2$
(c) $(2mn)^2 + (n^2− m^2)^2 = (n^2 + m^2)^2$
(a) We proceed as follows
$$ \begin{align} (3n)^2 + (4n)^2 & = (3^2 + 4^2)n^2 \\ \\ & = 5^2n^2 \\ \\ & = (5n)^2 \end{align} $$
(b) We proceed as follows
$$ \begin{align} (2n)^2 + (n^2− 1)^2 & = 4n^2 + (n^2)^2 + 1 - 2n^2 \\ \\ & = 2n^2 + (n^2)^2 + 1 \\ \\ & = (n^2 + 1)^2 \end{align} $$
(c) We proceed as follows
$$ \begin{align} (2mn)^2 + (n^2− m^2)^2 & = 4m^2n^2 + n^4 + m^4 -2n^2m^2 \\ \\ & = 2m^2n^2 + n^4 + m^4 \\ \\ & = (n^2 + m^2)^2 \end{align} $$