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Factoring the Sum of Squares

revised 21 Apr 2011

Copyright © 2009–2013 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.

Contents:

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.

Sophie Germain’s Identity

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

x4 + 4y4 = (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.

Example 1:  Factor 4x4 + 625y4.

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

4x4 + 625y4 = (2x² + 25y² + 10xy) (2x² + 25y² − 10xy)

Higher Powers

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

Example 2:  (u³)² + (v³)² = u6 + v6. Can anything be done?

The answer is yes. As you may know, An+Bn can be factored on the reals if n is an odd integer:

An+Bn = (A+B) (An-1 − An-2B + An-3B2 ... − ABn-2 + Bn-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 u6+v6 as the sum of two cubes:

u6+v6 = (u²)³+(v²)³ = (u²+v²) ( (u²)² − u²v² + (v²)² ) = (u²+v²) (u4 − u²v² + v4)

Example 3: x10+1024y10 is a sum of squares, (x5)2 + (32y5)2. But it’s also a sum of fifth powers, x10+1024y10 = (x2)5 + (4y2)5

Use the above factorization for the sum of fifth powers:

A5+B5 = (A+B) (A4 − A3B + A2B2 − AB3 + B4)

x10+1024y10 = (x2)5 + (4y2)5

    =(x2+4y2) ( (x2)4 − (x2)3(4y2) + (x2)2(4y2)2 − (x2)(4y2)3 + (4y2)4 )

    =(x2+4y2) ( x8 − 4x6y2 + 16x4y4 − 64x2y6 + 256y8)

Complex Numbers

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 − (iB)² = (A+iB) (A−iB)

Example 4:  x²+16 = (x+4i) (x−4i)

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

Example 6:  16x4+81y4 = (4x²)² + (9y²)² = (4x²+9iy²) (4x²−9iy²)

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 9iy² as (3 √i y)² = (3 ((1+i)/√2) y)² = ((3+3i)y/√2)², so you have a sum of squares and a difference of squares:

16x4+81y4 = [4x²+9iy²] [4x²−9iy²]

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

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

Why are there no new is 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+3i)y/√2 ] [ 2x − i(3+3i)y/√2 ] [ 2x + (3+3i)y/√2 ] [ 2x − (3+3i)y/√2 ]

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

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

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

16x4+81y4

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

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

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

16x4+81y4

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


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