(In reply to
Algebraic approach by Steve Herman)
I think this approach may work.
We seek (a^2+1)(b^2+1)(c^2+1)(d^2+1)(e^2+1)
which expands to the sums
{5}(abcde)^2 +
{4}[(abcd)^2 etc. 5 four way products] +
{3}[(abc)^2 etc. 10 three way products] +
{2}[(ab)^2 etc. 10 two way products] +
{1}[a^2 + b^2 + c^2 + d^2 + e^2]
{0} 1
and we seek {5}+{4}+{3}+{2}+{1}+{0}
We know as Steve pointed out
[1] (a+b+c+d+e) = 0
[2] (ab + ac + ... cd +ce + de) = 0 [sum of 10 pairwise products]
[3] (abc + abd + ... + bce + bde + cde) = 1 [sum of 10 three way products]
[4] (bcde + acde + abde + abce + abcd) = 0
[5] (abcde) = -1
So we have
{0}=1
[5]=1={5}
[1]^2 = {1}+[2] = 0
so {1} = 0
I don't have time to finish right now. But this approach shows promise.
This only leaves {2}, {3}, and {4}
|
|
Posted by Jer
on 2021-05-09 12:01:59 |
re: Algebraic approach
