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.

• Any odd number is the difference of two (consecutive) squares. $2n+1=(n+1)^2-n^2$. May 17 at 15:20
• Yes @AndreasBlass any odd number is the difference of two consecutive squares. However, in this case the differences are among non-consecutive squares (except the first one). Hence requesting for a proof / counter example. May 18 at 11:33
• My comment is a proof, unless you change the question to require non-consecutive $m$ and $n$. May 18 at 12:49
• If you really want non-consecutive $m$ and $n$, then just observe that any odd number that is divisible by $3$ is necessarily of the form $6n+3$ and therefore the difference of non-consecutive squares $(n+2)^2-(n-1)^2$. May 18 at 14:46
• @KarriChandrasekhar to prove by induction, you have to use: $(a^2 - b^2)(c^2 - d^2) = (ac + bd)^2 - (ad + bc)^2$. But of course you have to prove that these two are not consecutive.
– sku
Jun 15 at 21:50

Note that any product of two distinct odd numbers or two distinct even numbers can be written as a difference of squares. Let $$p be both odd or both even. Then $$(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.

• The other answer is correct, of course... fixing $p=1$ shows that any odd number is the difference of consecutive squares. For non-consecutive squares, you only need to stipulate $p>1$, which of course is still possible in OP's case. May 18 at 13:51

You might note that the difference sequence of the sequence of squares is exactly the odd numbers. Here's the sequence of squares:

$$0, 1, 4, 9, 16, 25, 36, \ldots$$

Here's the sequence of the differences:

$$1, 3, 5, 7, 9, 11, \ldots$$

That means every odd number is the difference of two consecutive squares. And your number is odd.

• Yes B.Goddard, any odd number is the difference of two consecutive squares. However, in this case the differences are among non-consecutive squares (except the first one). Hence requesting for a proof / counter example. May 18 at 11:33
• @KarriChandrasekhar Then maybe you should re-word your question. You say nothing about "non-consecutive." So as it stands, your objection makes no sense. May 18 at 12:05