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 is a prime iff , where .

**Solution**: Note that if is even, then the expression is clearly composite. If is odd, say, for some , then we have

And, if , then both the factors above are greater than , and hence the expression is composite. Moreover, if , we have , which is prime. And, we are done.

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February 21, 2008 at 7:34 pm

John smithproblem 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 smithi mean is that a typing error? (you can delete this)

February 27, 2008 at 6:30 am

AnonymousThere 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 TrimblePerhaps 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 LamaTodd,

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 TrimbleYes, and I guess you know the quaternionic analogue of that identity? It tells you that if and are the sums of

foursquares, 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

eightsquares 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 LamaWell, I knew that as

Euler’s four square identitybut 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!