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i remember when problem like this i was solving in junior class.how old are you?
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3^3 is not 9
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One counterexample disproves it nuff nuff my stuff.
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if n^3 is odd, then n is odd. therefore a larger odd minus a smaller odd equals an even number, then even number plus one is odd.
if n^3 is even, then n is even. therefore an even minus an even number equals an even number, then plus one is odd.
no n exists such that n^3-n+1 is even.
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little bit easier proof, takes about 2 seconds. just factor n^3-n = n(n+1)(n-1), we need to prove this is always even because we add 1 to it in the original problem. if n is even n(n+1)(n-1) is obviously even and n^3-n +1 odd. if n is odd then n(n+1)(n-1) is even since both n+1 and n-1 are both even. therefore n^3-n is always even and n^3-n+1 is always odd, hence never even.
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wow...this is really cool stuff!!!
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I found a easier way, just test every integer, worked for me.
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If n is even, n cubed is even because any number times an even number is even.
n^3 - n would also be even, as an even number subtracted from an even number is even.
Answer is odd because even number + 1 is odd.
If n is odd, n cubed is odd because an odd number multiplied by successive odd numbers is still odd.
However, n^3 - n is even because an odd number subtracted from an odd number is an even number.
Again, answer is odd because even + 1 is odd.
Q.E.D.
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bro...
bro...
just use induction to prove that it's odd
Stupidly long route... n^3 - n + 1 = n(n-1)(n+1) + 1. And n(n-1)(n+1) is a product of 3 consecutive numbers, and at least one of these must be even, thus n(n-1)(n+1) is even, and adding the 1 gives the result.
rohypnol55 2 years ago 22
you totally need to get a bigger whiteboard
LogInForPaper 2 years ago 14