factoring

• Jul 21st 2010, 11:03 AM
anthonye
factoring
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.
• Jul 21st 2010, 11:24 AM
HallsofIvy
Remember that factoring allows you to solve equations: \$\displaystyle a^2- b^2= 0\$ can be solved by \$\displaystyle (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 \$\displaystyle a^2+ b^2\$ with real coefficients (we can factor it as \$\displaystyle 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 \$\displaystyle 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 \$\displaystyle a^2+ b^2\$ must be positive, not 0.
• Jul 21st 2010, 02:05 PM
anthonye
thanks hallsofivy i also read something about there being no common factors can you help on this matter
• Jul 21st 2010, 11:34 PM
HallsofIvy
I have no idea what you mean by that. Are you still referring to \$\displaystyle a^2+ b^2\$?

\$\displaystyle a^2- b^2\$ also has "no common factors" but can be factored.
• Jul 22nd 2010, 12:51 AM
anthonye
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.
• Jul 22nd 2010, 03:35 AM
HallsofIvy
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 \$\displaystyle a^2- b^2\$ does NOT have an obvious "common factor", we could rewrite it as \$\displaystyle 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: \$\displaystyle a(a- b)+ b(a- b)\$. And now we can see that the two terms have a "common factor" of a-b: \$\displaystyle (a+ b)(a- b)\$. But that is not always the simplest thing to do- for something like \$\displaystyle a^2- b^2\$ it is simplest to remember the simple formula, \$\displaystyle a^2- b^2= (a- b)(a+ b)\$.

Notice that if we try to do the same thing with \$\displaystyle a^2+ b^2\$ we would have \$\displaystyle 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 \$\displaystyle a^2+ b^2\$ or, more generally, \$\displaystyle a^{2n}+ b^{2n}\$ is what I said before: \$\displaystyle 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.