Existence of a natural number X

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Homework Statement


Given M \in N, show that there exists an X \in N such that for all n \geq X, n^2+n+1 \succ M



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The Attempt at a Solution


Since both M and X are natural numbers and I am just trying to prove the existence of a certain natural number X, I thought that i could just set X = M.
Then, n^2+n+1 \geq X^2+X+1 since n \geq X.
And X^2+X+1=M^2+M+1\succM.
So, n^2+n+1\succM.
Is this a sufficient proof for the existence of X?
It just doesn't feel like a full proof, should X be more limited?
 
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If that is the correct statement of the problem, I don't see anything wrong with your argument. But I have an uneasy feeling like you do. Since it seems so trivial I wonder if the original problem is mis-copied or misunderstood.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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