# Here's a problem

• Dec 28th 2009, 09:52 AM
wonderboy1953
Here's a problem
I give you a bigrade: $\displaystyle a^n + b^n + c^n = d^n + e^n + f^n; n = 1,2$

Now give me ab + ac + bc = de + df + ef (please show all work) and find values for a - f that makes the equation work.
• Dec 28th 2009, 09:26 PM
CaptainBlack
Quote:

Originally Posted by wonderboy1953
I give you a bigrade: $\displaystyle a^n + b^n + c^n = d^n + e^n + f^n; n = 1,2$

Now give me ab + ac + bc = de + df + ef (please show all work) and find values for a - f that makes the equation work.

Do you know how to solve this? If not it does not belong in this forum

CB
• Dec 28th 2009, 10:45 PM
simplependulum
Quote:

Originally Posted by wonderboy1953
I give you a bigrade: $\displaystyle a^n + b^n + c^n = d^n + e^n + f^n; n = 1,2$

Now give me ab + ac + bc = de + df + ef (please show all work) and find values for a - f that makes the equation work.

we now have two equations

$\displaystyle a + b + c = e + d + f$

$\displaystyle a^2 + b^2 + c^2 = e^2 + d^2 + f^2$

And do you know that

$\displaystyle \sum^n a_k^2 = \left ( \sum^n a_k \right )^2 - 2 \sum_{ i \neq j} a_i a_j$

$\displaystyle a^2 + b^2 + c^2 = ( a + b + c )^2 - 2(ab+ bc + ca)$

$\displaystyle d^2 + e^2 + f^2 = ( d + e + f )^2 - 2 (de + ef +fd )$

so we have $\displaystyle ( a + b + c )^2 - 2(ab+ bc + ca) = ( d + e + f )^2 - 2 (de + ef +fd )$

$\displaystyle ab + bc + ca = de + ef + fd$
• Dec 29th 2009, 09:10 AM
wonderboy1953
Response to Captain Black
Of course I know how to solve this problem, otherwise I wouldn't have posted it here (I didn't want to immediately post the solution as that would have given away the show).

simplependulum got the answer right. I would have squared both sides and then use simple algebra to derive the end result to show that it follows naturally - the same difference, but I think my way would be more understandable. Now for the next part (which I know how to do):

Starting from:

http://www.mathhelpforum.com/math-he...ae3726ca-1.gif, can you show through adding 1 to each and every member of this bigrade that you still have equality? [the first step is to derive $\displaystyle (a + 1)^2 + (b + 1)^2 + (c + 1)^2 = (d + 1)^2 + (e + 1)^2 + (f + 1)^2$]
• Dec 29th 2009, 09:24 AM
wonderboy1953
BTW
Here are one set of values (among the infinite) that make the bigrade work: a = 1, b = 6, c = 8, d = 2, e = 4 and f = 9. The reason why it's important to specify is you have can doubters (like Kronecker) who say that nothing is proven unless you can show everything.
• Jan 3rd 2010, 12:00 PM
wonderboy1953
Solving the puzzle
I left it to be proven that:

reduces to:

http://www.mathhelpforum.com/math-he...ae3726ca-1.gif

Opening parentheses, you get:

$\displaystyle a^2 + 2a + 1 + b^2 + 2b + 1 + c^2 + 2c + 1 = d^2 + 2d + 1 + e^2 + 2e + 1 + f^2 + 2f + 1$

Now rearranging terms and combining some:

$\displaystyle a^2 + b^2 + c^2 + 2(a + b + c) + 3 = d^2 + e^2 + f^2 + 2(d + e + f) + 3$

Obviously the 3 cancels out. Also, by assumption, since this is a bigrade,
a + b + c = d + e + f also cancels out from the equation (with the coefficient 2) which leaves you with:

$\displaystyle a^2 + b^2 + c^2 = d^2 + e^2 + f^2$

which is true by assumption. Here's my main point: with any multigrade, you can add any positive integer (to each and every term) and get a new multigrade, no exceptions.