Prove that if a rectangle's sides are all odd, then it's diagonal is irrational? In trying to write an alternate and simple proof that at least one leg of a right triangle is a multiple of 4 using Dickson's method of generating triples, I came across quite an interesting observation that ifall the sides of a rectangle is odd, then it's diagonal is irrational. For example, consider a rectangle of length 5 units and breadth 3 units. It's diagonal,by the Pythagorean theorem is 34^0.5, which is irrational. This is the same for lots of rectangles. Is there a general proof for this, or can this whole thing be disproved?

If the sides of the rectangle are integers, then the square of the length of the diagonal is also an integer since 
$d^2 = l^2 + w^2$
and the integers are closed under multiplication and addition.  
The only integers that have rational square square roots are perfect squares.  Since  
$d = \sqrt{l^2 + w^2}$
the length of the diagonal is irrational unless $d^2 = l^2 + w^2$ is a perfect square.
If an integer is a perfect square, then it is either the square of an even number or an odd number.  
If the diagonal is an even number, then $d = 2k$ for some $k \in \mathbb{N}$, so $d^2 = (2k)^2 = 4k^2$.
If the diagonal is an odd number, then $d = 2k - 1$ for some $k \in \mathbb{N}$, so $d^2 = (2k - 1)^2 = 4k^2 - 4k + 1 = 4(k^2 - k) + 1$.
Hence, the square of any integer will have remainder $0$ or $1$ when divided by $4$.
If both the length and width are odd numbers, then $l^2 + w^2$ has remainder $2$ when divided by $4$.  To see this, suppose that $l = 2m - 1$ and $w = 2n - 1$, where $m$ and $n$ are positive integers.  Then
\begin{align*}
d^2 & = l^2 + w^2\\
    & = (2m - 1)^2 + (2n - 1)^2\\
    & = 4m^2 - 4m + 1 + 4n^2 - 4n + 1\\
    & = 4(m^2 - m + n^2 - n) + 2
\end{align*}
Since $d^2 = l^2 + w^2$ has remainder $2$ when divided by $4$, it is not a perfect square.  Thus, $d = \sqrt{l^2 + w^2}$ is irrational.
What I have not shown is why the only integers that have rational square roots are perfect squares.  It hinges on unique factorization.  You can show that $n$ is a perfect square if and only if each prime in its prime factorization appears an even number of times.  For example, $$144 = 2^4 \cdot 3^2 = (2^2 \cdot 3)^2 = 12^2$$ while $12 = 2^2 \cdot 3$ is not a perfect square.  If $n$ is not a perfect square, then some prime in its factorization must appear an odd number of times.  Suppose $n$ is not a perfect square and $$\sqrt{n} = \frac{r}{s}$$ where $r, s$ are integers with $s \neq 0$.  Then 
\begin{align*}
n & = \frac{r^2}{s^2}\\
ns^2 & = r^2   
\end{align*}
If $p$ is a prime that appears in the prime factorization of $n$ an odd number of times, then its appears in the prime factorization of $ns^2$ an odd number of times, while it appears in the prime factorization of $r^2$ an even number of times, a contradiction of unique factorization.  
