factoring

**anthonye** · July 21st 2010, 11:03 AM

Hi all,
why can't a^2+b^2 be factored in fact non of the following factor

a^4+b^4
a^8+b^8
a^16+b^16

the factors are all even numbers and some even factors can be broken up to to produce a prime number factor ie 10 become 2^5 but i can't quite see why these sums can't be factored.

Thanks for the help.

**HallsofIvy** · July 21st 2010, 11:24 AM

Remember that factoring allows you to solve equations: $a^2- b^2= 0$ can be solved by $(a- b)(a+ b)= 0$ so that a- b= 0 or a= b and a+ b= 0 or a= -b are solutions.

If it were possible to factor $a^2+ b^2$ with real coefficients (we can factor it as $a^2+ b^2= a^2- (-b)^2= (a- bi)(a+ bi)$ ) then it would be possible to find non-zero a and b so that $a^2+ b^2= 0$ and of course, that is not true. Since the square of any non-zero number is positive, if a and b are not both 0, then $a^2+ b^2$ must be positive, not 0.

**anthonye** · July 21st 2010, 02:05 PM

thanks hallsofivy i also read something about there being no common factors can you help on this matter

**HallsofIvy** · July 21st 2010, 11:34 PM

I have no idea what you mean by that. Are you still referring to $a^2+ b^2$ ?

$a^2- b^2$ also has "no common factors" but can be factored.

**anthonye** · July 22nd 2010, 12:51 AM

Why can a^2-b^2 be factored and a^2+b^2 can't I thought there needed to be a common factor(s) to factor something.

**HallsofIvy** · July 22nd 2010, 03:35 AM

No, you do not have to have "common factors"- at least not obvious ones.

IF you have a "common factor", like the "a" in ab+ ac, then you can simply use the "distributive law": ab+ ac= a(b+ c). But with binomials or trinomials it can be a bit more complicated.

While $a^2- b^2$ does NOT have an obvious "common factor", we could rewrite it as $a^2- ab+ ab- b^2= (a^2- ab)+ (ab- b^2$ . Now the first binomial has a "common factor" of a and the second has a "common factor" of b: $a(a- b)+ b(a- b)$ . And now we can see that the two terms have a "common factor" of a-b: $(a+ b)(a- b)$ . But that is not always the simplest thing to do- for something like $a^2- b^2$ it is simplest to remember the simple formula, $a^2- b^2= (a- b)(a+ b)$ .

Notice that if we try to do the same thing with $a^2+ b^2$ we would have $a^2- b^2= a^2- ab+ ab+ b^2= a(a- b)+ b(a+ b)$ and now we cannot continue further.

But the real reason we cannot factor $a^2+ b^2$ or, more generally, $a^{2n}+ b^{2n}$ is what I said before: $a^{2n}+ b^{2n}= 0$ cannot have any real roots except a= b= 0. If we were able to factor, with real coefficients, we could find an infinite number of real number values for a and b that would satisfy the equation.

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