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Suppose that $a$ and $b \in \mathbb{Z}^+$ satisfy $a^2b|a^3+b^3$. Prove that $a=b$.

I have reduced the above formulation to these two cases. Assuming $b = a + k$. Proving that any of the below two implies that $a=b$ will be enough.

$$a^2b|(a+b)^3 - 3ab^2$$

$$a^2b|2a^3+3a(a+k)+k^3$$

I can't proceed from here. How should I proceed from here?

Thanks.

Nizar
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TheRandomGuy
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4 Answers4

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By hypothesis $\ n = \dfrac{a^3\!+b^3}{a^2b} = \dfrac{a}b + \left(\dfrac{b}a\right)^2\! =\, x+x^{-2}\,\overset{\large {\times\, x^2}}\Longrightarrow\,x^3-n\,x^2 + 1 = 0$

By the Rational Root Test $\ a/b\, =\, x\, = \pm 1\ \ $ QED

Generally applying RRT as above yields the degree $\,j+k\,$ homogeneous generalization

$$a,b,c_i\in\Bbb Z,\,\ a^{\large j}b^{\large k}\mid \color{#c00}1\:\! a^{\large j+k}\! + c_1 a^{\large j+k-1} b + \cdots + c_{\large j+k-1} a b^{\large j+k-1}\! + \color{#c00}1\:\!b^{\large j+k}\Rightarrow\, a = \pm b \qquad $$

$\qquad\qquad\ \ \ \ \ \ $ e.g. $\ a^2b \mid a^3 + c_1 a^2b + c_2 ab^2 + b^3\,\Rightarrow\, a = \pm b $

Alternatively the statement is homogeneous in $\,a,b\,$ so we can cancel $\,\gcd(a,b)^{\large j+k}$ to reduce to the case $\,a,b\,$ coprime. The dividend $\,c\,$ has form $\,a^{\large n}\!+b^{\large n}\! + abm\,$ so by Euclid it is coprime to $a,b$ thus $\,a,b\mid c\,\Rightarrow\, a,b = \pm1$.

Remark $\ $ The proof in lhf's answer is precisely the standard proof of the Rational Root Test specialized to this particular polynomial. The Rational Root Test concisely encapsulates all divisibility results of this (homogeneous) form.

Bill Dubuque
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  • (+1) I saw this question on the front page because of your edit. I had written essentially the same answer before actually opening the question. This is definitely the simplest approach I can think of. – robjohn Aug 10 '19 at 20:34
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Hint: Here, $a,b$ can be negative. It suffices to prove that $|a|=|b|$ when $\gcd(a,b)=1$ and $a^2b\mid a^3+b^3$.

Batominovski
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  • Isn't $gcd(a, b) = a = b$ and not necessarily $1$ if $a=b$? – Deepak Jun 15 '16 at 14:14
  • Which part of logic eluded you? – Batominovski Jun 15 '16 at 14:31
  • If $\gcd(a,b)=1$ and $a^2b\mid a^3+b^3$, then $|a|=|b|=1$. – user236182 Jun 15 '16 at 14:37
  • Sorry, I didn't work through the question, but if the only solutions have $a = b= 1$, then there is no issue. – Deepak Jun 15 '16 at 14:48
  • @Deepak One can eliminate the gcds and obtain a much simpler proof using the Rational Root Test - see my answer. – Bill Dubuque Jun 18 '16 at 19:55
  • @BillDubuque Nice solution. However, to say that the Rational Root Test doesn't involve gcd is most likely not true. I think that there is no proof of this theorem not involving gcd. – Batominovski Jun 18 '16 at 21:00
  • @Batominovski I meant explicit use of gcds. One could argue that most all proofs in elementary number theory ultimately depend on gcds (or prime factorizations). Such distinctions will become clearer when one studies factorization theory in more general domains (gcd, Bezout, PID, UFDs, etc). – Bill Dubuque Jun 18 '16 at 21:04
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Write $d=\gcd(a,b)$ and $a=dA$, $b=dB$, with $\gcd(A,B)=1$.

Then we can cancel $d^3$ on both sides of $a^2b \mid a^3+b^3$ and get $A^2B \mid A^3+B^3$.

This implies that $A^2 \mid B^3$ and $B \mid A^3$, and so $A=B=1$.

Indeed, $A^2 \mid B^3$ implies that every prime that divides $A$ divides $B$. Since $\gcd(A,B)=1$, we must have $A=1$. But then $B \mid A^3$ implies $B=1$.

lhf
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Let $a=da_1$, $b=db_1$, where $d=\gcd(a,b)$, $a_1,b_1\in\mathbb Z^+$. Then $\gcd(a_1,b_1)=1$, $\gcd\left(a_1^2,b\right)=1$ and $$a^2b\mid a^3+b^3$$

$$\iff d^3a_1^2b_1\mid d^3a_1^3+d^3b_1^3$$

$$\iff a_1^2b_1\mid a_1^3+b_1^3$$

$$\iff \begin{cases}a_1^2\mid a_1^3+b_1^3\\b_1\mid a_1^3+b_1^3\end{cases}\iff \begin{cases}a_1^2\mid b_1^3\\ b_1\mid a_1^3\end{cases}$$

Since $\gcd\left(a_1^2,b_1^3\right)=\gcd\left(b_1,a_1^3\right)=1$ and $a_1,b_1\in\mathbb Z^+$, we get $a_1=b_1=1$. Therefore $a=b=\gcd(a,b)$.

user236182
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