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Prove or disprove:
For any integer n, the number n² + n + 1 is odd.
 one year ago
 one year ago
Prove or disprove: For any integer n, the number n² + n + 1 is odd.
 one year ago
 one year ago

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JoãoVitorMCBest ResponseYou've already chosen the best response.0
i think that to but i have to prove some how
 one year ago

AriPottaBest ResponseYou've already chosen the best response.1
5^2 + 5 +1 25 + 5 +1 = 31 (odd) 1^2 + 1 + 1 1 + 1 +1 = 3 (odd)
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
break it up into cases Case 1: If n is even,... Case 2: If n is odd,...
 one year ago

AriPottaBest ResponseYou've already chosen the best response.1
i can't find any number that disproves it...
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
AriPotta you proved it works for those particular values, but you didn't prove it for all integers
 one year ago

AriPottaBest ResponseYou've already chosen the best response.1
2^2 + 2 + 1 4 + 2 + 1 = 7 (odd)
 one year ago

JoãoVitorMCBest ResponseYou've already chosen the best response.0
ok i'll try using cases
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
Case 1: n is an even integer Let k be any integer, so 2k is always even. Let n = 2k n^2 + n + 1 (2k)^2 + (2k) + 1 4k^2 + 2k + 1 2(2k^2 + k) + 1 2q + 1 ... Let q = 2k^2 + k (you can easily show that q is always an integer) So if n = 2k (ie if n is an even integer), then n^2 + n + 1 is equivalent to 2q + 1, which is an odd integer. I'll let you do case 2
 one year ago

benzillaBest ResponseYou've already chosen the best response.0
An even + an odd is an odd . an odd + odd is an even number, and an even + even is an even number. If n is odd n^2 is odd so we have an odd + odd + 1 which is the same as odd + even so it will be odd. If n is an even n^2 is even so n^2 +n will be even. If you add 1 to an even number it is odd. Aripotta has the right idea. Testing numbers works for odd and even in all cases.
 one year ago

JoãoVitorMCBest ResponseYou've already chosen the best response.0
thank you all of you!
 one year ago
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