# Show that there isn't $k\in\Bbb Z^+$ such that $k^3 + 2k^2 + 2k + 1$ is cube number

We need to show that there doesn't exist positive integer $$k$$ such that $$k^3 + 2k^2 + 2k + 1$$ is a cube number.

I tried solving it in this way: Firstly let the cube number we are looking for be $$l$$, we can write $$l = k + d$$, then it must hold $$k^3 + 2k^2 + 2k+1=(k+d)^3$$. If we extend the binom $$(k+d)$$ on third power $$k^3+2k^2+2k+1 = k^3+3k^2d+3kd^2+d^3$$. In order for this to be true it must be that $$2 = 3d, 2=3d^2 \text{ and }d^3=1$$. Surely this is contradiction, since such $$d$$ doesn't exist.

Is my work good enough or are there any mistakes that make it wrong, are there easier ways to show this?

• With $k=-1$, we have $k^3+2k^2+2k+1=-1+2-2+1=0=0^3$. With $k=0$, we have $0^3+0+0+1=1=1^3$. – Hagen von Eitzen Oct 19 at 10:45
• I just edited the question, I forgot to mention that $k$ must be positive. – someone123123 Oct 19 at 11:00

Hint :

For all $$K \in \mathbb N,$$

$$k^3 + 2k^2 + 2k + 1 \gt(k)^3$$

And

$$k^3 + 2k^2 + 2k + 1 \lt k^3 + 3k^2 + 3k + 1 \implies k^3 + 2k^2 + 2k + 1 \lt (k+1)^3$$

• We only care about positive integer $k$ – someone123123 Oct 19 at 10:58
• @someone123123 you should specify this in the question, – The Demonix _ Hermit Oct 19 at 10:59
• I'm sorry about this mistake, I will correct it now. – someone123123 Oct 19 at 10:59

Your argument is fallacious because an equation $$k^3+ak^2+bk+c=k^3+dk^2+ek+f$$ does not imply that coefficient-wise $$a=d$$, $$b=e$$, $$c=f$$. Such a conclusion would be valid only if the equation were known to hold for enough (three, in this case) distinct values of $$k$$.