Should be easy: $a|(2b+a)^2$ and $a$ is odd implies $a|b$?

Question

$a|(2b+a)^2$ and $a$ is odd implies $a|b$?

I would say that: $$ a|(2b+a)^2 $$ $$ a|4b^2+4ab+a^2 $$ $$ a|4b^2$$ $$ a|b^2$$ But I don't think that this means that $a|b$, does it? How can I prove that $a|b$, if it's possible? Thanks

Answer 1

No. If $a=9$ and $b=3$ we have $9|225 $ but not $9|3$.

Answer 2

You are correct, odd $\rm\:a\mid (2b+a)^2 \iff a\mid b^2.\:$ But $\rm\:a\mid b^2\Rightarrow\: a\mid b\:$ holds iff $\rm\:a\:$ is squarefree.

Answer 3

$a | (2b + a) ^2 = 4b^2 + 4ab + a^2 \Longleftrightarrow a | 4b^2 \Longleftrightarrow a | b^2$

There are an infinite number of counterexamples: all are given by any odd $b$ (but $1$), and $a$ any odd divisor of $b^2$ but not of $b$ (could be $b^2$).

Answer 4

No, it doesn't imply $a | b$.

The statement isn't correct anyway: take $a = 9$ and $b = 6$.

Then $9 | (2*6 + 9)^2 = 441$ but $9$ doesn't divide $6$.

Answer 5

The claim is not true for $a=9$ and $b=3$.

Edit: I can say something about how I arrived at this counterexample. One sees that $a|(2b+a)^2$ if $a|b^2$. So I just tried some small numbers such that $a|b^2$, but $a \nmid b$.