Wikipedia:Reference desk/Archives/Mathematics/2006 November 23

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November 23

Quick Math Question

I have this problem for homework, could someone point me in the right way, or help me set up the equation? Find three consecutive positive odd integers such that the sum of the squares of the first two is 15 less than the square of the third. Answer: 1,3,5, or 3,5,7. This unit is on Quadratics. Jamesino 02:11, 23 November 2006 (UTC)

I'd start by calling the smallest of the three numbers 'x', figuring out what to call the other two in terms of 'x', and then translating the condition you're given into an equation. -GTBacchus^(talk) 02:19, 23 November 2006 (UTC)

x^2 + (x+2)^2 +15 = (x+4)^2 That takes care of the consecutive integers, but how do i guarantee them to be odd? Jamesino 02:22, 23 November 2006 (UTC)

Well, as long as the roots of the quadratic turn out to be integers, the conditions of the problem guarantee that they'll be odd. If they were even, all their squares would be even, and the sum of two of them couldn't differ from the third by 15. If you want to specify that a number is odd, you can always call it '2k+1' where 'k' is an integer, but when you set up a quadratic, the solution(s) obtained may not even be rational. Clearly, the problem was written by someone who already knew the solutions would be odd integers. I'd just solve the quadratic and then say "oh, aren't we lucky, the solutions really are odd integers". -GTBacchus^(talk) 02:29, 23 November 2006 (UTC)

Ah, ok Thanks. Jamesino 02:49, 23 November 2006 (UTC)

Here are some pieces to use:

• We can write any odd integer as 2k+1, where k is any integer.
• If n is an odd integer, then n+2 and n+4 are the next two after it.
• If m is an integer, then m² is its square.

When we put these together according to the problem statement, we create a quadratic equation in k. Massive simplification reduces it to standard three term form. We find that both its roots give suitable integer solutions. Done. --KSmrq^T 04:16, 23 November 2006 (UTC)

Algebra stumper

Years ago, a calculus professor gave me this one. The problem is to either solve the following set of equations by giving x, y and z; or to prove no such solution exists (among the complex numbers)

$x+y+z=1\,\!$

$x^{2}+y^{2}+z^{2}=2\,\!$

$x^{3}+y^{3}+z^{3}=3\,\!$

—The preceding unsigned comment was added by Plf515 (talk • contribs) 03:50, 2006 November 24.

Suppose a is one of x, y, and z. Then $(a-x)(a-y)(a-z)=0$, so

$a^{3}-(x+y+z)a^{2}+(xy+xz+yz)a-(xyz)a=0\,\!$

We already know one of the coefficients, and we can find the other two:

$(x+y+z)^{2}=1\,\!$

$x^{2}+y^{2}+z^{2}+2xy+2xz+2yz=1\,\!$

$2+2(xy+xz+yz)=1\,\!$

$xy+xz+yz=-1/2\,\!$

$(x+y+z)(x^{2}+y^{2}+z^{2})=2\,\!$

$x^{3}+y^{3}+z^{3}+x^{2}y+x^{2}z+xy^{2}+xz^{2}+y^{2}z+yz^{2}=2\,\!$

$(x+y+z)(xy+xz+yz)=-1/2\,\!$

$3xyz+x^{2}y+x^{2}z+xy^{2}+xz^{2}+y^{2}z+yz^{2}=-1/2\,\!$

$x^{3}+y^{3}+z^{3}-3xyz=5/2\,\!$

$3-3xyz=5/2\,\!$

$xyz=1/6\,\!$

So, x, y, and z are the roots of the cubic polynomial $a^{3}-a^{2}-(1/2)a-1/6$, which are about 1.43, 0.215 + 0.265i, and 0.215 - 0.265i. —Keenan Pepper 05:30, 24 November 2006 (UTC)

We might explicitly point out that all three equations lead to symmetric polynomials in x, y, and z. This has a double relevance. First, as shown, it assists in finding one solution. Second, it tells us that any permutation of one solution is also a solution.

A modern tool, that may not have been available when the problem was first posed, is the computation of a Gröbner basis for the three polynomials using Buchberger's algorithm. One such basis here is

${\begin{aligned}&6z^{3}-6z^{2}-3z-1,\\&2y^{2}+2zy-2y+2z^{2}-2z-1,\\&x+y+z-1.\end{aligned}}$

The cubic in z should look familiar. --KSmrq^T 06:52, 24 November 2006 (UTC)

Expressed algebraically, the real root of the 3rd degree polynomial equals:

${\frac {w}{w^{2}-w-1}},{\mbox{ where }}w={\sqrt[{3}]{5+{\sqrt {26}}}}.$

This was found by first using the substitution $z:=1/\zeta$ and subjecting the result to Cardano's third degree technique of interrogation.  --Lambiam^Talk 09:14, 24 November 2006 (UTC)

Thanks~ Plf515 10:56, 24 November 2006 (UTC)plf515

Thanks for mentioning the Gröbner Basis solution as well. – b_jonas 17:50, 25 November 2006 (UTC)