# Factoring the Sum of Squares

revised 20 Jul 2015

Copyright © 2009–2015 by Stan Brown, Oak Road Systems

revised 20 Jul 2015

Copyright © 2009–2015 by Stan Brown, Oak Road Systems

**Summary:**
The commonest algebra mistake is probably rewriting
A²+B² as (A+B)². “You can’t factor the sum
of two squares on the reals!” your teacher tells you. While
that’s generally true, there are some interesting
exceptions.

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Since Euler’s time at least, generations of students have
tried to “factor” A²+B² as (A+B)². The lure
of this siren song is so strong that I see even calculus students
commit this blunder.
Generations of teachers have sighed despairingly and tried to
get students to remember that
**a sum of two squares can’t be factored on the reals**.

When explaining Solving Polynomial Equations I myself made that bare statement. But in correspondence in September 2009, Steve Schwartzman convinced me that I should say more. He proposed a sentence or two, but on the principle that a thing worth doing is worth overdoing, ...

It’s true that you can’t factor A²+B²
on the reals if A and B are just simple
variables, but if A and B have **internal structure** then the expression
may be factorable after all, if you can find some other pattern.
So it’s still true that a sum of squares
can’t be factored *as* a sum of squares on the reals.

This page looks at some of the cases where a sum of squares can be factored using other techniques.

The counterexample that Steve Schwartzman sent me in September 2009 is, as he told me, a form of Sophie Germain’s identity:

x^{4} + 4y^{4} =
(x² + 2y² + 2xy)
(x² + 2y² − 2xy)

Can you generalize this to a class of factorable sums of squares? Yes, you can.

Notice that the factors have the form of (P+Q)(P−Q),
which of course multiplies to P²−Q². This suggests
that, for factoring A²+B², it might be fruitful to look at
(A+B)² minus something. That’s all well and good, but minus
*what*?

The key is that (A+B)² = A²+2AB+B². Comparing that to A²+B², you see that there’s an extra term of 2AB. So you have

A² + B² = (A+B)² − 2AB

That right-hand side is factorable as a difference of squares,
*if* 2AB is a perfect square. And that’s our
factorization:

A² + B² = (A + B + √(2AB)) (A + B − √(2AB))

This identity is always true, but it’s useful for
factoring only **when 2AB is a perfect square**.

More specifically, 2AB must be a perfect square if
you want your factors to have
*rational coefficients*. If you allow non-rational factors, you
can factor more sums of squares, and if you allow
complex factors you can factor *any* sum
of squares.

**Example 1**: Factor 4x^{4} + 625y^{4}.

Solution: Let A = 2x² and B = 25y²; then 2AB = 100x²y² is a perfect square and √(2AB) = 10xy.

4x^{4} + 625y^{4} =
(2x² + 25y² + 10xy) (2x² + 25y² −
10xy)

If A and B are themselves odd powers, you can use a common pattern to factor A²+B².

**Example 2**: (u³)² + (v³)² =
u^{6} + v^{6}. Can anything be done?

The answer is yes.
As you may know, A^{n}+B^{n} can be factored on
the reals if n is an odd integer:

A^{n}+B^{n} =
(A+B) (A^{n-1} − A^{n-2}B +
A^{n-3}B^{2} ... − AB^{n-2} +
B^{n-1}) for n odd

You’ve probably learned the simplest case, n = 3:

A³+B³ = (A+B) (A² − AB + B²)

So the solution is to rewrite u^{6}+v^{6} as
the sum of two cubes:

u^{6}+v^{6} =
(u²)³+(v²)³ =
(u²+v²) ( (u²)² − u²v² +
(v²)² ) =
(u²+v²) (u^{4} − u²v² + v^{4})

**Example 3**: x^{10}+1024y^{10} is a sum of squares,
(x^{5})^{2} + (32y^{5})^{2}.
But it’s also a sum of fifth powers,
x^{10}+1024y^{10} =
(x^{2})^{5} + (4y^{2})^{5}

Use the above factorization for the sum of fifth powers:

A^{5}+B^{5} =
(A+B) (A^{4} − A^{3}B +
A^{2}B^{2} − AB^{3} +
B^{4})

x^{10}+1024y^{10} =
(x^{2})^{5} + (4y^{2})^{5}

=(x^{2}+4y^{2})
( (x^{2})^{4} −
(x^{2})^{3}(4y^{2}) +
(x^{2})^{2}(4y^{2})^{2} −
(x^{2})(4y^{2})^{3} +
(4y^{2})^{4} )

=(x^{2}+4y^{2})
( x^{8} −
4x^{6}y^{2} +
16x^{4}y^{4} −
64x^{2}y^{6} +
256y^{8})

Usually when you factor, you are looking for real factors. But if you allow complex factors then you can always factor A²+B², like this:

A² + B² =
A² − (−1·B²) =
A² − (*i*B)² =
(A+*i*B) (A−*i*B)

**Example 4**: x²+16 = (x+4*i*) (x−4*i*)

**Example 5**: 25p²+49q² =
(5p)² + (7q)² =
(5p+7*i*q) (5p−7*i*q)

**Example 6**: 16x^{4}+81y^{4} =
(4x²)² + (9y²)² =
(4x²+9*i*y²) (4x²−9*i*y²)

This one’s interesting because you can go further.
(If not interested, feel free to skip this.)
If only you can write *i* as a square—in other words,
if you can find the square
root of *i*—then the two factors become a
sum of squares and a difference of squares.
√*i* is covered in some trig classes: the
principal square root is (1+*i*)/√2, and the other square
root is minus that.

Therefore, you can rewrite 9*i*y² as
(3 √i y)² =
(3 ((1+*i*)/√2) y)² =
((3+3*i*)y/√2)², so you have a sum of squares and a
difference of squares:

16x^{4}+81y^{4} =
[4x²+9*i*y²] [4x²−9*i*y²]

=
[ (2x)² + ((3+3*i*)y/√2)² ]
[ (2x)² − ((3+3*i*)y/√2)² ]

=
[ 2x + *i*(3+3*i*)y/√2 ]
[ 2x − *i*(3+3*i*)y/√2 ]
[ 2x + (3+3*i*)y/√2 ]
[ 2x − (3+3*i*)y/√2 ]

Why are there no new *i*s in the second pair of factors?
Because that’s good old A²−B² =
(A+B)(A−B). The first pair of factors can be simplified a
bit:

[ 2x + *i*(3+3*i*)y/√2 ]
[ 2x − *i*(3+3*i*)y/√2 ]
[ 2x + (3+3*i*)y/√2 ]
[ 2x − (3+3*i*)y/√2 ]

=
[ 2x + (3*i*−3)y/√2 ]
[ 2x − (3*i*−3)y/√2 ]
[ 2x + (3+3*i*)y/√2 ]
[ 2x − (3+3*i*)y/√2 ]

=
[ 2x − (3−3*i*)y/√2 ]
[ 2x + (3−3*i*)y/√2 ]
[ 2x + (3+3*i*)y/√2 ]
[ 2x − (3+3*i*)y/√2 ]

You can factor out the 1/√2 and get rid of the inner parentheses:

16x^{4}+81y^{4}

=
(1/√2)
[2√2 x − (3−3*i*)y]
[2√2 x + (3−3*i*)y]
[2√2 x + (3+3*i*)y]
[2√2 x − (3+3*i*)y]

=
(1/√2)
(2√2 x − 3y + 3*i*y)
(2√2 x + 3y − 3*i*y)
(2√2 x + 3y + 3*i*y)
(2√2 x − 3y − 3*i*y)

And finally you can reorder the four trinomials to show the combinations of plus and minus signs:

16x^{4}+81y^{4}

=
(1/√2)
(2√2 x + 3y + 3*i*y)
(2√2 x + 3y − 3*i*y)
(2√2 x − 3y + 3*i*y)
(2√2 x − 3y − 3*i*y)

**20 Jul 2015**: Pete Hepple pointed out that 2²+3² = (2+3+√12)(2+3−√12), even though 2·2·3 = 12 isn’t a perfect square. I realized I had nowhere made it explicit that by “factoring” I meant “factoring on the rationals”, so I added that.- (intervening changes suppressed)
**13 Sep 2009**: New document, inspired by correspondence with Steve Schwartzman.

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