## shubhamsrg Group Title 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.. one year ago one year ago

1. hartnn Group Title

simple for 2nd degree...but donno how to generalize.. u got for 2nd degree ?

2. sauravshakya Group Title

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)

3. shubhamsrg Group Title

yep i got it for 2 nd degree,,same problem,,how to generalize @hartnn 2nd step ? @sauravshakya

4. sauravshakya Group Title

But I dont this will lead to the solution or not: p_0=1 p_1=a+b+c...k

5. experimentX Group Title

try making use of complex conjugates.

6. experimentX Group Title

though not sure if this will work.

7. sauravshakya Group Title

(x-a)(x-b)(x-c)...(x-k) =x^n-(a+b+c+...+k)x^n-1 +...+(abc...k)

8. sauravshakya Group Title

Actually its: p_1=-(a+b+c...k)

9. experimentX Group Title

$(x - ia_1)(x-ia_2)(x-ia_3) .... (x-ia_n) = x^n + i(p_1)(x^{n-}1) - (p_2)(x^{n-2})-ip_3 x^{n-3} .... \\ (p_n-1)(x) + p_n$

10. experimentX Group Title

multiply those ... and put back x=1

11. shubhamsrg Group Title

can euler's formulla help ?

12. experimentX Group Title

$(x + ia_1)(x+ia_2)(x+ia_3) .... (x+ia_n) = x^n - i(p_1)(x^{n-}1) + (p_2)(x^{n-2})+ip_3 x^{n-3} .... \\ (p_n-1)(x) + p_n$ looks like awful generalization.

13. experimentX Group Title

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

14. experimentX Group Title

looks promising though looks rigorous.

15. experimentX Group Title

on the putting x=1, you would naturally get (1+ a^2) (1 +b^2)....(1 +k^2) on LHS side.

16. experimentX Group Title

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 $\Large\sum_{j=0}^n \sum_{k=0}^n p_j p_k (-1)^k (i)^{j+k} (1)^j$

17. shubhamsrg Group Title

i'd give this a try will come back to you @experimentX

18. experimentX Group Title

yeah sure ... still looks pretty rigorous.

19. experimentX Group Title

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.

20. siddhantsharan Group Title

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.

21. siddhantsharan Group Title

Is it fine guys?

22. experimentX Group Title

yeah seem simpler that way.

23. mukushla Group Title

im late ... lol nice :)

24. shubhamsrg Group Title

cool! B| thanks a lot @siddhantsharan