HINT: $a^2+b^2=c^2$ cannot hold for $a,b$ odd and $c$ even. (using congruences) Here is what I have: Suppose the contrary. Thus,
(2k+1)(2k+1) + (2j+1)(2j+1) = (2p)(2p)
Take mod 2 of both sides
[1][1]+[1][1]=[0]
[2]=[0]
[0]=[0]
No contradiction ... am I approaching this correctly? I want to figure this out myself but would like a small hint

With $a$, $b$ odd we may write
$a = 2m + 1, \tag 1$
$b = 2n + 1; \tag 2$
then
$a^2 = 4m^2 + 4m + 1, \tag 3$
$b^2 = 4n^2 + 4n + 1, \tag 4$
$a^2 + b^2 = 4(m^2 + n^2 + m + n) + 2; \tag 5$
also,
$c = 2p, \tag 6$
whence
$c^2 = 4p^2; \tag 7$
if
$a^2 + b^2 = c^2, \tag 8$
then substituting in (3), (4) and (7) we find
$4(m^2 + n^2 + m + n) + 2 = 4p^2; \tag 9$
we reduce this modulo $4$ and obtain
$2 \equiv 0 \mod 4, \tag{10}$
a contradiction which shows there are no such $a$, $b$, and $c$.
