Those who love (elementary) problem-solving eventually come across the Sophie Germain identity. It has lots of applications in elementary number theory, algebra and so on. The identity states
.
Indeed, note that
More can read about Marie-Sophie Germain, a brilliant mathematician, here and here.
Let us use the above identity in solving a couple of problems. Here is the first one.
Problem 1: Evaluate 
Solution: A first glance tells us that the sum should somehow “telescope.” But the denominator looks somewhat nasty! And, it is here that the above identity comes to our rescue. Using the Sophie Germain identity, we first note that
We thus have
Here is another one.
Problem 2: Show that 
Solution: Note that if 
And, if 
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February 21, 2008 at 7:34 pm
John smith
problem 2) if n=1 it is obvious no? do you mean if n is a natural number?
February 21, 2008 at 7:37 pm
John smith
i mean is that a typing error? (you can delete this)
February 27, 2008 at 6:30 am
Anonymous
There are no typing errors. “iff” is convention for “if and only if”. So while it is obvious if n=1, as you said, you have to prove the “only if” part.
May 1, 2008 at 7:55 pm
Todd Trimble
Perhaps this is well-known to readers out there, but there’s a way of figuring out the Sophie Germain factoring of
even if you didn’t know it before, using complex numbers.
Since this polynomial is homogeneous (all monomial terms have the same degree), the problem can be reduced to one variable by putting
and then factoring 
; the original polynomial can be reinstated as 
.
The roots of
are the complex fourth roots of 
. Now one can use De Moivre’s theorem, which tells us that the 
complex roots of 
are the 
distinct values 
for 
.
So the fourth roots of
are:
A short calculation yields their values:
. Now group these into pairs of conjugates; for example, 
and 
form a conjugate pair. We then get factors with real coefficients:
multiplying each by
, we get the two factors in the Sophie Germain identity.
May 1, 2008 at 8:22 pm
Vishal Lama
Todd,
I hadn’t seen that nice proof before, but now that you have shown it, I very much like it! It does remind me of another (“complex number” ) proof of a different identity:
. Here we set 
and 
and then use the fact 
.
Well, in this case, we don’t invoke De Moivre’s theorem, but then the above proof is similar in spirit to the one you showed in proving the Sophie Germain identity.
May 2, 2008 at 1:46 pm
Todd Trimble
Yes, and I guess you know the quaternionic analogue of that identity? It tells you that if
and 
are the sums of four squares, then so is 
.
Quaternions form a 4-dimensional vector space over the reals, with basis elements
which are multiplied according to
If you fiddle around with this for a while, you find
Using these rules, you can then work out the product of two quaternions
, 
(using distributivity: multiplication by a quaternion should be an 
-linear transformation). It is noteworthy that quaternionic multiplication is associative but noncommutative.
Also, as for the complex numbers, there is a conjugation operation
and using the rules above, one sees that
for two quaternions
[it suffices to check this just on the basis elements]. Also one finds that for a quaternion 
, we have
a sum of four squares, and this defines the square norm of a quaternion,
. As a result,
where the last equation follows since quaternions commute with real scalars like
. This immediately implies that formal sums of four squares are closed under multiplication! with the explicit rule for that given by quaternionic multiplication. Using this, one can prove that every positive integer is the sum of four squares, by showing first that every prime is the sum of four squares.
There is even an octonionic analogue of that which implies that formal sums of eight squares are closed under multiplication! [But the buck stops there -- there's no analogue for higher powers of 2.] John Baez has written extensively about these things (e.g., here), and there is a nice book by Conway and Smith, On Quaternions and Octonions, which has a wealth of information about these things.
May 2, 2008 at 3:58 pm
Vishal Lama
Well, I knew that as Euler’s four square identity but had not seen the “quaternionic” proof of it! I only knew a somewhat messy proof that involved using Lagrange’s identity. So, thanks for showing the above proof!