menpo wrote: Positive integers $a,b,c$ satisfy the equations $a^2=b^3+ab$ and $c^3=a+b+c$. Prove that $a=bc$. Let $d=\gcd(a,b)$ and $a=da'$ and $b=db'$ with $\gcd(a',b')=1$ First equation becomes $a'^2=db'^2+a'b'$ and so $b'|a'^2$ and so $b'=1$ and equation becomes $d=a'^2-a'$ and so $a=a'd=a'^3-a'^2$ and $b=db'=d=a'^2-a'$ Plugging $(a,b,c)=(a'^3-a'^2,a'^2-a',c)$ in second equation, we get $c^3=a'^3-a'+c$, which is $(c-a')(c^2+ca'+a'^2-1)=0$ And so, since all are positive, $c=a'$ and $\boxed{a=da'=bc}$ Q.E.D.

pco wrote: menpo wrote: Positive integers $a,b,c$ satisfy the equations $a^2=b^3+ab$ and $c^3=a+b+c$. Prove that $a=bc$. Let $d=\gcd(a,b)$ and $a=da'$ and $b=db'$ with $\gcd(a',b')=1$ First equation becomes $a'^2=db'^2+a'b'$ and so $b'|a'^2$ and so $b'=1$ and equation becomes $d=a'^2-a'$ and so $a=a'd=a'^3-a'^2$ and $b=db'=d=a'^2-a'$ Plugging $(a,b,c)=(a'^3-a'^2,a'^2-a',c)$ in second equation, we get $c^3=a'^3-a'+c$, which is $(c-a')(c^2+ca'+a'^2-1)=0$ And so, since all are positive, $c=a'$ and $\boxed{a=da'=bc}$ Q.E.D. Good. The problem is still true, when $a,b,c$ are positive reals.

menpo wrote: Positive integers $a,b,c$ satisfy the equations $a^2=b^3+ab$ and $c^3=a+b+c$. Prove that $a=bc$. menpo wrote: The problem is still true, when $a,b,c$ are positive reals. Considering $a,b,c$ as positive reals : Plugging $b=c^3-a-c$ in first equation, we get : $a\in(0,c^3-c)$ (and so $c>1$) and $a^3-a^2(3c^3-3c-2)+a(c^3-c)(3c^3-3c-1)-(c^3-c)^3=0$ Which is $(a-(c^3-c^2))(a^2-a(c+1)(2c^2-c-2)+(c^2+c)(c^2-1)^2)=0$ and so Either $a=c^3-c^2$ and so $b=c^3-a-c=c^2-c$ and indeed $a=bc$ Either $a^2-a(c+1)(2c^2-c-2)+(c^2+c)(c^2-1)^2=0$ This quadratic has discriminant $(c+1)^2(4-3c^2)$ and so we need $c^2\le\frac 43$ Product of roots of the quadratic is $>0$ and so we need sum of roots positive (in order to have $a>0$) and so $2c^2-c-2>0$ And this is impossible since then $c<2c^2-2\le \frac 83-2<1$ And so unique solution $(a,b,c)=(c^3-c^2,c^2-c,c)$ where $c>1$ and indeed $a=bc$

pco wrote: menpo wrote: Positive integers $a,b,c$ satisfy the equations $a^2=b^3+ab$ and $c^3=a+b+c$. Prove that $a=bc$. menpo wrote: The problem is still true, when $a,b,c$ are positive reals. Considering $a,b,c$ as positive reals : And so unique solution $(a,b,c)=(c^3-c^2,c^2-c,c)$ where $c>1$ and indeed $a=bc$ Nice. Here is the short solution: Take $a = bk,$ by first equation $k > 1,$ then by substituting we obtain $c^3 - c = k^3 - k \implies (c-k)(c^2 + ck + k^2) = (c-k),$ but $c^2+ck+k^2 > 1$, so $k = c$.

Consider the first equation as a quadratic in a. Then $a = \frac{b \pm b\sqrt{4b+1}}{2}$. Because a is positive and $\sqrt{4b+1} \geq 1$, $a = \frac{b + b\sqrt{4b+1}}{2}$. Also, $c \geq 1$, so $c^3-c-a-b$ is strictly increasing and has only one solution. It suffices to show that $c = \frac{a}{b} = \frac{1+\sqrt{4b+1}}{2}$ is a solution, which is easily verified.

We first prove that $b\mid a$. Suppose for the sake of contradiction there exists a prime $p$ such that $\nu_p(b)>\nu_p(a)$. Then \[ 2\nu_p(a)=\nu_p(b)+\nu_p(a+b^2)=\nu_p(b)+\nu_p(a), \]a contradiction. Note that $\frac{a}{b}=1+\frac{b^2}{a}$. For a prime $p$ such that $\nu_p(b^2)>\nu_p(a)$, $1+\frac{b^2}{a}$ is relatively prime with $p$ so $\nu_p(a)=\nu_p(b)$. Since $\frac{b^2}{a}$ is an integer, we always have $2\nu_p(b)=\nu_p(b^2)\geq\nu_p(a)$. Thus we may write $a=m^2n$ and $b=mn$ for relatively prime positive integers $m$ and $n$. Now \[ m=\frac{a}{b}=1+\frac{b^2}{a}=1+n \]so $c^3-c=(n+1)^2n+(n+1)n=(n+1)^3-(n+1)$. Since $x\mapsto x^3-x$ is increasing for $x\geq 1$ it follows that $c=n+1=m$, as desired. $\square$

Notice that since the first equation is a quadratic in $a$ we can use the quadratic formula $$a=\frac{b\pm \sqrt{b^2+4b^3}}{2}$$Clearly the $\pm$ must be a $+$. Also for $\sqrt{b^2+4b^3}=b\sqrt{1+4b}$ to be a square we must have $b=x(x+1)$ for some positive integer $x$. Then we get that $a=x(x+1)^2$. So $c^3-c=(c-1)(c)(c+1)=x(x+1)+x(x+1)^2=x(x+1)(x+2)$. Thus $c=x+1$ and the result follows.

Prove that b divides a and write a=bk,then b=k(k+l). c(c²-1)=k(k²-1) and then bound the values of k to prove k=c

\begin{align*} a^2 &= b^3 + ab \\ b &= ka \\ b^3 &= k^3a^3 \\ a^2 &= ka^3 + ka^2 \\ ka &= \frac{1}{1-k} \\ a &= \frac{1-k}{k} \\ k &= \frac{1}{1+a} \\ b &= \frac{a}{1+a}\\ c^3 &= a + \frac{a}{1+a} + c \\ c(c+1)(c-1) &= \frac{a^2 + 2a}{a+1} = \text{Integer} \\ a+1 &\mid a^2 + 2a \\ \text{Now, } (a+1)^2 &= a^2 + 2a + 1 \\ \text{Hence, } a+1 \mid 1 \implies a = 0 \\ \therefore b &= 0 \\ \therefore c &= 0, 1, -1 \\ \text{Hence, } a &= bc \end{align*}