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pratu043
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Find the condition that x^n + y^n may be divisible by x + y.
 one year ago
 one year ago
pratu043 Group Title
Find the condition that x^n + y^n may be divisible by x + y.
 one year ago
 one year ago

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sauravshakya Group TitleBest ResponseYou've already chosen the best response.2
IF n is odd
 one year ago

pratu043 Group TitleBest ResponseYou've already chosen the best response.0
How do you prove that?
 one year ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
sum of odd roots factor always factor out (x+y)
 one year ago

pratu043 Group TitleBest ResponseYou've already chosen the best response.0
What do you mean by that?
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.2
I know how, PLZ wait sec trying to write it down
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.2
Oh got it.... let f(x)=x^n + y^n
 one year ago

pratu043 Group TitleBest ResponseYou've already chosen the best response.0
So you should use factor theorem for this?
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.2
now, when f(x) is divided by (x+y) then REMAINDER = f(y) = (y)^n + y^n
 one year ago

pratu043 Group TitleBest ResponseYou've already chosen the best response.0
ok got that.
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.2
Now, for x^n + y^n to be divisible by x + y.. remainder = 0 or, (y)^n + y^n=0
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.2
which is only possible when n is odd
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.2
got it?
 one year ago

pratu043 Group TitleBest ResponseYou've already chosen the best response.0
ok so when n is odd, (y)^n = y so (y)^n + y^n = y + y = 0 thanks!!
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.2
oh its like this: when n is odd, (y)^n = (y)^n so (y)^n + y^n = (y)^n + y^n = 0
 one year ago
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