Is there a solution to $a^2+b^2+c^2+d^2+e^2=f^2$ with a, b, c, d, e, and f all odd integers? Examining integer solutions to $a^2+b^2+c^2+d^2+e^2=f^2$ it would seem there are only two cases. Either (1) $a, b, c, d, e,$ and $f$ are all even or (2) $f$ and exactly one of the summands is even and the rest are odd. Is it possible to have all odd integers as a solution set? If not, why not?
 A: The square of an odd integer is always of the form $8k+1$.  Therefore, the sum of the squares of five odd integers is of the form $8k+5$ and thus cannot be a perfect square.
The square of an even integer is either of the form $8k$ or $8k+4$.  So they could all be even or four odd and one of the form $8k$ with a sum of the form $8k+4$.  That's all you've got for cases.

ETA: As Barry Cipra notes, four even numbers that add to $8k$ and an odd number summing to an odd number is also technically and actually possible.
A: To give a relatively complete answer, there are three possibilities for the parity of integer solutions to $a^2+b^2+c^2+d^2+e^2=f^2$: All five variables on the LHS are even (and so is $f$), exactly one variable on the LHS is odd (and so is $f$), and exactly one variable on the LHS is even (and so is $f$). If we restrict to primitive solutions, where $\gcd(a,b,c,d,e)=1$, then only the latter two cases occur. Trivial examples are
$$0+0+0+0+1=1$$
and
$$1+1+1+1+0=4$$
Slightly less trivial examples, with positive integers, are
$$4+4+4+4+9=25$$
and
$$9+9+9+9+64=100$$
If you want examples with distinct positive numbers, here's a pair:
$$4+16+36+64+1=121$$
and
$$1+9+25+49+16=100$$
The existence of infinitely many primitive solutions is guaranteed by the sum-of-four-squares theorem applied to $f^2-1$.  There's probably a simpler proof with an explicit family, but nothing comes immediately to mind. (But see the "Added later" below.) 
A proof that there are no solutions with either two or three odd variables on the LHS can be given by considering things mod $4$; the proof that there are no solutions with all five variables odd requires an argument mod $8$, as in Matthew Daly's answer.
Added later: I was being blind. Let $a$, $b$, $c$, and $d$ be any four numbers such that $S=a^2+b^2+c^2+d^2$ is odd, and let $gh$ be any factorization of $S$, such as $1\cdot S$. Then solve $g=f-e$, $h=f+e$ for $e$ and $f$, i.e., $e=(h-g)/2$, $f=(h+g)/2$. Alternatively, let $a$, $b$, $c$, and $d$ be odd, which implies $S\equiv4$ mod $8$, and take a factorization of $S$ with both factors even. In particular, if $S=4N$, then we get
$$a^2+b^2+c^2+d^2+(N-1)^2=(N+1)^2$$
Note that $S=4N\equiv4$ mod $8$ implies $N$ is odd, as we've seen it must be (i.e. $N-1$ must be even). Since we're free to take $a$, $b$, $c$, and $d$ to be any odd numbers, it's clear we can produce arbitrarily many primitive solutions with distinct positive numbers.
