if a,b,c...k are the roots of some n degree equation, x^n + (p_1)(x^n-1) + (p_2)(x^n-2)....(p_n-1)(x) + p_n = 0 p_i would mean " i " is in the subscript .. then how to prove that : (1+ a^2) (1 +b^2)....(1 +k^2) = (1 - p_2 + p_4 -p_6 ....)^2 + (p_1 - p_3 + p_5 - .........)^2 hope i could write it clearly..

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if a,b,c...k are the roots of some n degree equation, x^n + (p_1)(x^n-1) + (p_2)(x^n-2)....(p_n-1)(x) + p_n = 0 p_i would mean " i " is in the subscript .. then how to prove that : (1+ a^2) (1 +b^2)....(1 +k^2) = (1 - p_2 + p_4 -p_6 ....)^2 + (p_1 - p_3 + p_5 - .........)^2 hope i could write it clearly..

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simple for 2nd degree...but donno how to generalize.. u got for 2nd degree ?
The first step would be this one: x^n + (p_1)(x^n-1) + (p_2)(x^n-2)....(p_n-1)(x) + p_n =(x-a)(x-b)(x-c)...(x-k)
yep i got it for 2 nd degree,,same problem,,how to generalize @hartnn 2nd step ? @sauravshakya

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But I dont this will lead to the solution or not: p_0=1 p_1=a+b+c...k
try making use of complex conjugates.
though not sure if this will work.
(x-a)(x-b)(x-c)...(x-k) =x^n-(a+b+c+...+k)x^n-1 +...+(abc...k)
Actually its: p_1=-(a+b+c...k)
\[ (x - ia_1)(x-ia_2)(x-ia_3) .... (x-ia_n) = x^n + i(p_1)(x^{n-}1) - (p_2)(x^{n-2})-ip_3 x^{n-3} .... \\ (p_n-1)(x) + p_n\]
multiply those ... and put back x=1
can euler's formulla help ?
\[ (x + ia_1)(x+ia_2)(x+ia_3) .... (x+ia_n) = x^n - i(p_1)(x^{n-}1) + (p_2)(x^{n-2})+ip_3 x^{n-3} .... \\ (p_n-1)(x) + p_n \] looks like awful generalization.
looks like the general term would be \[ \huge \prod_{i=1}^{n}(x-ia_0) = \sum_{j=0}^n x^{n-j} p_j i^j\\ \huge \prod_{i=1}^{n}(x+ia_0) = \sum_{k=0}^n x^{n-k} p_k (-i)^k\] where p_0 = 1 try multiplying those two and put x=1
looks promising though looks rigorous.
on the putting x=1, you would naturally get (1+ a^2) (1 +b^2)....(1 +k^2) on LHS side.
on the right side you would be \[ \Large\sum_{j=0}^n \sum_{k=0}^n p_j p_k (-1)^k (i)^{j+k} \] the imaginary terms would cancel out because it only happens if j+k is odd ... so either j is odd or k is odd. ... if that's the case then the other variable will be even and we have \[ \Large\sum_{j=0}^n \sum_{k=0}^n p_j p_k (-1)^k (i)^{j+k} (1)^j\]
i'd give this a try will come back to you @experimentX
yeah sure ... still looks pretty rigorous.
the other way would be \[ (x+a^2)(x+b^2) ... = x^n + (a^2+b^2 ..)x^{n-1} + ... \\ x^n + ((a+b+...)^2 - 2(ab +bc +...))x^{n-1} + ... \\ x^n + (-p_1^2 - 2p_2)x^{n-1} + ... \] this looks equally rigorous.
Its pretty simple. Write P(x) = (x-a)(x-b)......(x-k) = x^n + (p_1)(x^n-1) + (p_2)(x^n-2)....(p_n-1)(x) + p Put=> x = i in this identity. => (i - a)(i-b).........(i-k) = (1 - p_2 + p_4 -p_6 ....)i + (p_1 - p_3 + p_5 - .........) OR it can also be: => (i - a)(i-b).........(i-k) = (1 - p_2 + p_4 -p_6 ....) + (p_1 - p_3 + p_5 - .........)i In either case take modulus both sides to prove the required.
Is it fine guys?
yeah seem simpler that way.
im late ... lol nice :)
cool! B| thanks a lot @siddhantsharan

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