<s>[INST] How to prove the product of first n consecutive odd numbers is a square less than another square? I have observed for first few values of consecutive odd numbers, the result is always of the form:
$m^2 - n^2$, where $m$ and $n$ are two distinct positive integers. That is:
$$1\cdot 3\cdot 5 \cdot 7\cdots (2k-1) = m^2 - n^2$$
For example:
$
1\cdot 3 = 3 = 2^2 - 1^2 \\
1\cdot 3\cdot 5 = 15 = 4^2 - 1^2 \\
1\cdot 3\cdot 5\cdot 7 = 105 = 11^2 - 4^2 \\
1\cdot 3\cdot 5\cdot 7\cdot 9 = 945 = 31^2 - 4^2 \\
\vdots
$
But not sure, how to prove it.
Here is an attempt using induction:
Let it be true for some value of k, that is:
$1\cdot 3\cdot 5\cdot 7\cdot 9\cdots(2k - 1) = m^2 - n^2$
Then when $k$ takes the value of $k+1$, we have
$$\begin{align}
1\cdot 3\cdot 5\cdot 7\cdot 9\cdots(2k - 1)\cdot(2k + 1)
&= (m^2 - n^2)\cdot(2k + 1)\\
&= (m^2 - n^2)\cdot {(k+1)^2 - k^2}
\end{align}$$
and got stuck here.
Can you please suggest to proceed further or an altogether a different way of proving so or prove me wrong.
Thanks in advance.
 [/INST] Note that any product of two distinct odd numbers or two distinct even numbers can be written as a difference of squares. Let $p<q$ be both odd or both even.  Then $(q-p)/2$ and $(q+p)/2$ are distinct integers, and
$$
((q+p)/2)^2 - ((q-p)/2)^2=(q^2/4+pq/2+p^2/4)-(q^2/4-pq/2+p^2/4)=pq.
$$
So this is a fairly weak condition, and one that the product of consecutive odd numbers certainly meets.
</s>