Is it possible to element \$a^2+b^2+c^2\$ ? If we make this right into only 2 factors, I understand it has to look choose this:

\$(a+b+c+cdot cdot cdot )(a+b+c+cdot cdot cdot )\$ . Yet I don"t know exactly how to eliminate the \$2(ab+bc+ac)\$ yet I have actually no idea what else should go in the parenthesis. How have the right to you figure out exactly how to variable this? Thanks. No. I will take into consideration the situation of factorizing over the complicated numbers, or any sub-field (e.g. The real numbers). Many of my discussion will no really rely on the choice of field.

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Any factorization (into polynomials) must be into polynomials of degree \$1\$, i.e.\$\$ a^2 + b^2 + c^2 overset?= (alpha_1 a + eta_1 b + gamma_1 c + delta_1)(alpha_2 a + eta_2 b + gamma_2 c + delta_2) \$\$where \$alpha_1,dots,delta_2\$ are facility numbers. If friend don"t psychic doing some work, girlfriend can increase this out. Friend might as well rescale \$alpha_1 leadsto alpha_1" = 1\$, \$eta_1 leadsto eta_1" = eta_1 / alpha_1\$, \$gamma_1 leadsto gamma_1" = gamma_1 / alpha_1\$, \$delta_1 leadsto delta_1" = delta_1 / alpha_1\$, \$alpha_2 leadsto alpha_2" = alpha_1alpha_2\$, ..., \$delta_2 leadsto delta_2" = delta_2alpha_2\$. This allows you assume the \$alpha_1 = 1\$, and you instantly conclude \$alpha_2 = 1\$. Currently looking at the coefficients of \$ab\$, you view that \$eta_1 = -eta_2\$, and on the various other hand the coefficient the \$eta^2\$ provides \$eta_1eta_2 = 1\$; from this equations you deserve to conclude that \$eta_1 = pm i\$, and also \$eta_2\$ is its negation. A comparable argument looking at the coefficients of \$ac\$ and also \$c^2\$ provides \$gamma_1 = pm i\$, and also \$gamma_2\$ is the negation. Yet this leads to a contradiction as soon as looking at the coefficient of \$bc\$.

Let me mention an important trick when searching for factorizations of highly symmetric things. That is a (nontrivial) theorem the factorization the polynomials (in any number of variables) is unique in the adhering to sense: if you have actually two factorizations the the very same polynomial, such the each aspect itself has actually no factors, climate the two factorizations agree approximately reordering the list and multiplying state on the perform by nonzero (complex, say) numbers.

So, let"s assume we have actually a factorization, say the one above. We can try to generate other factorizations. Well, the left hand next is real, in the feeling of being invariant under complicated conjugation. Therefore the complicated conjugate that the appropriate hand side is additionally a factorization. It complies with that we have either one of two possibilities:

Possibility 1: The lists are not reordered. Therefore, over there is a nonzero complicated number \$eta\$ such the \$aralpha_1 = eta alpha_1\$, \$ar eta_1 = etaeta_1\$, \$ar gamma_1 = etagamma_1\$, \$ar delta_1 = etadelta_1\$, \$aralpha_2 = eta^-1 alpha_2\$, \$ar eta_2 = eta^-1eta_2\$, \$ar gamma_2 = eta^-1gamma_2\$, \$ar delta_2 = eta^-1delta_2\$. It follows that \$areta = eta^-1\$. Upon multiply the first factor by \$sqrteta\$ and also the 2nd by \$sqrteta^-1\$, we deserve to assume without loss of generality that \$eta = 1\$. Yet then both determinants are real, and this is impossible: the LHS has no zeros except for \$alpha = eta = gamma = 0\$, vice versa, each factor on the right has actually a two-dimensional room worth that zeros.

Possibility 2: Under complicated conjugation, the 2 terms switch places. Again, we deserve to multiply by a constant, and also thereby assume that \$alpha_2 = ar alpha_1\$, \$eta_2 = ar eta_1\$, etc.

What other symmetries room there? Well, the left-hand side is invariant under every of \$a mapsto -a\$, \$b mapsto -b\$, and also \$cmapsto -c\$. Approximately some casework, this provides linear relations between the coefficients. The left-hand next is likewise invariant under convert \$a\$ through \$b\$, say.

Using every one of these symmetries significantly reduces the possible space of factorizations, as any factorization have to respect every symmetries as much as some scalar factors.

Here"s one final way to think about factorization problems. Let"s solve a non-zero value for \$c\$, and consider the room of remedies \$(a,b)\$ to \$0 = a^2 + b^2 + c^2\$ (\$a\$ and also \$b\$ might be complex numbers). If \$c^2\$ were negative and us were in search of real solutions, this would certainly be a circle.Your ability to visualize in \$mmsanotherstage2019.combb C^2\$ might not be great, however I assure you the the room of services is quiet something favor a circle. In particular, it is not a direct subspace of \$mmsanotherstage2019.combb C^2\$.

On the various other hand, the room of solutions \$(a,b)\$ to any kind of equation that the form \$0 = p(a,b)q(a,b)\$ whereby \$p,q\$ room degree-\$1\$ is a union of 2 (usually intersecting) straight subspaces in \$mmsanotherstage2019.combb C^2\$.

This type of argument additionally shows that \$a^2 + b^2 + c^2\$ has actually no factorizations over the complex numbers.

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Finally, a commenter said looking in ~ the quaternions. The quaternions have lots of actions that are various from commutative fields like \$mmsanotherstage2019.combb C\$. Indeed, end the quaternions, us have\$\$ (ai + bj + ck)(-ai - bj - ck) = a^2 + b^2 + c^2 \$\$provided \$a,b,c\$ are actual (and therefore commute with everything). If \$a,b,c\$ are permitted also to it is in quaternionic, then ns don"t see any type of convenient factorization, however I don"t have a proof that it is impossible.