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1

This is a big cheating But $(n+1)^3 = n^3+(n-1)^3$ is in contradiction with Fermat's Last Theorem


4

$n^3-6n^2-2=0 \implies n^2(n-6)=2$ $n<6 \implies n^2(n-6)<0$ $n=6 \implies n^2(n-6)=0$ $n>6 \implies n^2(n-6)\geq49$ Therefore, $\forall{n}\in\mathbb{N}:n^2(n-6)\neq2$


1

The graph of $f(x)=x^3-6x^2-2$ has a zero occurring at a non-integer value.


3

Now, you just have to show that the cubic $x^3-6x^2-2 = 0$ has no positive integer roots. By the rational root theorem, the only possible rational roots are $x = \pm 1, \pm 2$. Hence, the only possible positive integer roots are $x = 1,2$. Is $x = 1$ or $x = 2$ a root? If not, then there are no positive integer solutions $x = n$ to $x^3-6x^2-2 = 0$. ...


0

Before throwing around the pigeonhole principle, just try to reason through a few simple cases. It may help to look at the situation with sets of 4 or 5 elements, and then generalize what you find. Here are some things to think about: For #3, note that it is impossible to have distinct $a, b \in X$ such that $f(a) = f(b)$. So what does that tell you about ...



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