We have $c(c^2-c+1)=abX$, so $a=x_1y_1,\,b=x_2y_2$, where $x_1$ and $x_2$ are divisors of $c$; $y_1$ and $y_2$ are divisors of $c^2-c+1$. Let $x_2=c/x_3,\,y_2=(c^2-c+1)/y_3$. Then $a+b=x_1y_1+\frac{c(c^2-c+1)}{x_3y_3}$ is divisible by $c^2+1$, hence $x_1y_1+\frac{1}{x_3y_3}$ is divisible by $M:=c^2+1$ (because $c(c^2-c+1)\equiv 1$ modulo $M$). So, $x_1x_3y_1y_3+1=kM$. Note that $x_1x_3y_1y_3$ is a divisor of $c^2(c^2-c+1)^2$, so we may write $c^2(c^2-c+1)^2=(kM-1)x$. Considering both sides modulo $M$ we have $x\equiv -1$, so $x=lM-1$. Note that $k_0=1,\,l_0=c^2-2c+2$ or viceversa satisfy this equation, and it corresponds to $\{a,b\}=\{c,c^2-c+1\}$. Then, $(kM-1)(lM-1)=(M-1)(l_0M-1)$, from here $klM-(k+l)=l_0M-(l_0+1)$. We see that $k+l\equiv l_0+1$ modulo $M$. If $k+l=l_0+1$, then $kl=l_0$ and $\{k,l\}=\{1,l_0\}$. If not, $k+l\ge l_0+1+M$. We have $klM\ge (k+l-1)M$, so $l_0M-(l_0+1)=klM-(k+l)\ge (k+l-1)M-(k+l)=(k+l)(M-1)-M\ge (l_0+1+M)(M-1)-M$, it is a contradiction.

a and b are the roots of X^2 - (a+b)X + ab replace a and b by the relations given and chek that the discriminant must be positive because it's the same polynomial.This leads immediately to the result.

senouy: I'm afraid that it donesn't trivialy comes from the problem. You have no equations, but only divisibilities there....

Since the plynomial X^2 + (a+b)X + ab has two roots a and b Check that the numbers c and c^2 - c +1 LIES btween a and b (plug them in the polynomial and verify that we got two negative numbers by using the divisibility relation.And this is an impossible thing if one of the a or b i sdifferent from one of the c's.

Can you please post your complete solution ? The hints you are giving are clear enough

In fact c and c^2 - c + 1 lies betwwen a and b but i am wrong in saying that this is impossible unless a and b equal th c's.

What do you mean by sets of coincidence?

By symmetry, we can assume that $a \leq b$. We will analyze two situations: $i) \: a \leq c \: \Longrightarrow$ Since $c^2+1 \mid a+b \; , \; a+b \geq c^2+1$ and $b \geq c^2-c+1$ Let $c-a=x \; , \; b-(c^2-c+1)=y$. Since $a>0$ , then $x<c$ and $x,y \geq 0$ $c^2+1 \mid y-x$ Let $y=x+k(c^2+1)$ where $k \in \mathbb{Z}$ $b \mid c(c^2-c+1) \; \Longrightarrow \; c^2-c+1+y \mid cy$ $y \neq 0 \; \Longrightarrow \; cy \geq c^2-c+1+y \; \Longleftrightarrow \; c \neq 1\; , \; y \geq c+ \frac{1}{c-1}>c$ $\Longrightarrow \; y>x$ because $x \geq y \; \Longrightarrow \; c>x \geq y >c$ $y > x \; \Longrightarrow \; k>0$ $c^2-c+1+y \mid cy \; \Longrightarrow \; \frac{c[x+k(c^2+1)]}{(k+1)c^2-c+k+1+x} \in \mathbb{Z}$ $\Longrightarrow \; \frac{kc^2+xc}{(k+1)c^2-c+k+x+1} \in \mathbb{Z}$ $y \neq 0 \; \Longrightarrow \; k>0$ $\Longrightarrow \; kc^2+xc \geq (k+1)c^2-c+k+x+1 \; \Longrightarrow \; x(c-1) \geq c^2-c+k+1$ $\Longrightarrow \; c \neq 1 \; , \; x \geq c+\frac{k+1}{c-1}>c$. It is a contradiction since $c>x$. $y=0 \; \Longrightarrow \; b=c^2-c+1 \; , \; a \mid c \; , \; c^2+1 \mid c-a$ It is easy to show that the last one holds if and only if $c=a$. $ii) \; a>c \; \Longrightarrow$ $b<c^2-c+1$. Let $a=c+x \; , \; b=c^2-c+1-y$ $\Longrightarrow \; x,y >0 \; , \; c^2+1 \mid x-y$ $y>x \; \Longrightarrow \; y>c^2+1=b+c+y$ which is a contradiction. $x \geq y \; \Longrightarrow \; a \geq c+y \; , \; (c+y)(c^2-c+1-y)>c(c^2-c+1)$ $\Longleftrightarrow \; (c-1)^2>y \; \Longleftrightarrow \; b>c $ $\Longrightarrow \; ab>c(c^2-c+1)$ because $b \leq c \; \Longrightarrow \; a \leq b \leq c$ which is a contradiction. $ab>c(c^2-c+1)$ is also a contradiction. Solution is done.

erdos wrote: Let $a,b$ and $c$ be positive integers such that $ab$ divides $c(c^{2}-c+1)$ and $a+b$ is divisible by $c^{2}+1$. Prove that the sets $\{a,b\}$ and $\{c,c^{2}-c+1\}$ coincide. here is hint to another process: here $gcd(c,c^2-c+1)=1$ so $ab|c^2-c+1$ or $ab|c$,but $c^2+1|a+b$ so $a+b\ge c^2+1$

mathmdmb wrote: here $gcd(c,c^2-c+1)=1$ so $ab|c^2-c+1$ or $ab|c$ Sorry if I`m wrong but just put $a=c=3 , b=7$ !!! We have $ab \nmid c^2-c+1$ and $ab \nmid c$ .

$a=c$ is to be proved and here is a step to show that $a|c$ and $b|c^2-c+1$

mathmdmb wrote: $a=c$ is to be proved and here is a step to show that $a|c$ and $b|c^2-c+1$ Sorry , Maybe I was really bad in describe the wrong point . You said if $gcd(A,B)=1$ and $XY \mid AB$ then $XY \mid A$ or $XY \mid B$ . ($A,B,X,Y$ are natural.) And it`s not true.

mahanmath wrote: mathmdmb wrote: $a=c$ is to be proved and here is a step to show that $a|c$ and $b|c^2-c+1$ Sorry , Maybe I was really bad in describe the wrong point . You said if $gcd(A,B)=1$ and $XY \mid AB$ then $XY \mid A$ or $XY \mid B$ . ($A,B,X,Y$ are natural.) And it`s not true. that`s i pointed out,sorry(me too) for bad language, here we have wlog, $a|c \implies c\ge a$ and $b|c^2-c+1$,so let $c^2-c+1=bk,c^2+1=bk+c|a+b$ Then $a+b\ge bk+c\implies a-c\ge b(k-1)$ but $c\ge a$,so $a=c$ and then $k=1,b=c^2-c+1$

mathmdmb wrote: mahanmath wrote: mathmdmb wrote: $a=c$ is to be proved and here is a step to show that $a|c$ and $b|c^2-c+1$ Sorry , Maybe I was really bad in describe the wrong point . You said if $gcd(A,B)=1$ and $XY \mid AB$ then $XY \mid A$ or $XY \mid B$ . ($A,B,X,Y$ are natural.) And it`s not true. that`s i pointed out,sorry(me too) for bad language, here we have wlog, $a|c \implies c\ge a$ and $b|c^2-c+1$,so let $c^2-c+1=bk,c^2+1=bk+c|a+b$ Then $a+b\ge bk+c\implies a-c\ge b(k-1)$ but $c\ge a$,so $a=c$ and then $k=1,b=c^2-c+1$ I think you still doesn't understand. You said that "wlog $a|c$", but it isn't necessary to be true. Actually we prove that wlog $a=c$ and hence $a \mid c$, but this proof is the main problem.

can you explain why we are wrong here?Since $gcd(c,c^2-c+1)=1$,$gcd(a,b)=1$ because otherwise $c$ and $c^2-c+1$ would share a common factor and so I assumed $a|c$,not $a=c$,and I took it for an obvious conclusion.We must have this.

mathmdmb wrote: can you explain why we are wrong here?Since $gcd(c,c^2-c+1)=1$,$gcd(a,b)=1$ because otherwise $c$ and $c^2-c+1$ would share a common factor and so I assumed $a|c$,not $a=c$,and I took it for an obvious conclusion.We must have this. Yes, I understand your idea, but, unfortunately your claim is not true. As mahanmath said, we can easily find $a,b,c$ which don't satisfy your claim. Let us take $c=10, a=35, b=26$ As you said, $gcd(a,b)=gcd(c,c^2-c+1)=1$ and $ a \nmid c \; , \; a\nmid c^2-c+1 \; , \; b \nmid c \; , \; b \nmid c^2-c+1$, but $ab=c(c^2-c+1)$ and hence $ab \mid c(c^2-c+1)$

Thanx for the explanation

Hopefully this is right... I think this is basically what senouy suggested above. Let $c(c^2-c+1)=mab$ and $a+b=n(c^2+1)$. Then \begin{align*} (a-b)^2 =n^2(c^2+1)^2-\frac{4}{m}c(c^2-c+1) &=n^2c^4-\frac{4}{m}c^3+\left(2n^2+\frac{4}{m}\right)c^2-\frac{4}{m}c+n^2\\ &\ge n^2c^4-\frac{4}{m}c^3+\left(2n^2+\frac{4}{m^2n^2}\right)c^2-\frac{4}{m}c+n^2\\ &=\left(nc^2-\frac{2}{mn}c+n\right)^2. \end{align*}WLOG set $a\ge b$. Then \[a-b\ge nc^2-\frac{2}{mn}c+n\implies b\le\frac{1}{mn}c.\]But \[c^2+1|a+b\implies -mb^2\equiv mab=c(c^2-c+1)\equiv-c^2\pmod{c^2+1},\]so for some $k\ge0$, \[c^2+k(c^2+1)=mb^2\le m\left(\frac{1}{mn}c\right)^2=\frac{1}{mn^2}c^2.\]Clearly $k=0$ and $m=n=1$, so we're done.

erdos wrote: Let $a,b$ and $c$ be positive integers such that $ab$ divides $c(c^{2}-c+1)$ and $a+b$ is divisible by $c^{2}+1$. Prove that the sets $\{a,b\}$ and $\{c,c^{2}-c+1\}$ coincide. Set $abx=c(c^2-c+1)$ $(a+b)=(c^2+1)y$ ----We must prove that $x=y=1$ ---- Set $a+b=k$, $a-b=l$ obvious $k,l>0$ so transform to ... $(k^2-l^2)x=4c(c^2-c+1)$ (1) $k=(c^2+1)y$ (2) So ... $(c^2+1)^2y^2x-l^2x=4c(c^2-c+1)$ (3) But... $l^2x<k^2x=(c^2+1)^2y^2x=\frac{k^2}{k^2-l^2}*4c(c^2-c+1)<4c(c^2-c+1)$ So we prove that $l^2x<4c(c^2-c+1)$ (4) So from (3) , (4) we have... If $x,y>1$ then $(c^2+1)^2y^2x-l^2x>(c^2+1)^2y^2x-4c(c^2-c+1)>(c^2+1)^2-4c(c^2-c+1)$ $(c^2+1)^2-4c^2+4c-4>4c(c^2-c+1)$ ( for $c=1$ give $a=b=1$ so ok) So $x=y=1$ So $ab=c(c^2-c+1)$ , $a+b=c^2+1$ so ... $a^2-a(c^2+1)+c(c^2-c+1)=0$ , $D=(c-1)^4$ so $(a,b)=(c,c^2-c+1),(c^2-c+1,c)$ Δημήτρης

Define $a+b=l(c^2+1).$ Without loss of generality $a \le b \implies a \le \frac{(c^2+1)l}{2}.$ We have $$b=(c^2+1)l-a \mid c(c^2-c+1)=(c-1)(c^2+1)+1.$$$$\implies (c^2+1)l-a \mid (c-1)(c^2+1)l+l$$$$\implies (c^2+1)l-a \mid ((c-1)(c^2+1)l+l)-((c^2+1)l-a) \cdot (c-1)=(c-1)a+l.$$Then $$(c^2+1)l-a \le (c-1)a+l \implies a \ge cl.$$Now notice that $x((c^2+1)l-x)$ is strictly increasing in $(1,\frac{(c^2+1)l}{2})$, then $$c(c^2-c+1) \ge ab = a((c^2+1)l-a) \ge cl((c^2+1)l-cl)=l^2c(c^2-c+1)$$So $l=1$ and the equality occurs, i.e., $a=cl=c$ and $b=(c^2+1)l-a=c^2-c+1.$

Wlog assume that $a \le b$. There exists integer $k$ with $abk=c(c^2-c+1)$. $c^2+1|a+\frac{c(c^2-c+1)}{ak} \implies c^2+1|a^2k+c(c^2-c+1) \implies c^2+1|a^2k+1 \implies k \ge \frac{c^2}{a^2} \implies c(c^2-c+1) \ge \frac{bc^2}{a} \implies \frac{c^2-c+1}{c} \ge \frac{b}{a}$. Note that $c^2+1 \le a+b$. Thus $\frac{a+b-c}{c} \ge \frac{c^2-c+1}{c} \ge \frac{b}{a} \implies a \ge c \implies b \le c^2-c+1$ because of the first condition. Let $a=c+m, b=c^2-c+1-n$. First condition yields $m(c^2-c+1) \ge mn+cn ... (1)$, second condition yields $c^2+1|m-n ... (2)$, $a \le b \implies c^2-2c+1\ge m+n ... (3)$. If $m=n\neq 0$, we get $m \ge c^2-2c+1$ from $(1)$ and this contradicts with $(3)$. If $m \neq n$, we get $c^2-2c+1 \overset{(3)} \ge m+n \ge |m-n| \overset{(2)} \ge c^2+1$, contradiction. Thus the only possibility is $m=n=0$, sets $\{a,b\}$ and $\{c,c^{2}-c+1\}$ coincide

Let $a+b=k(c^2+1)$ and $c(c^2-c+1)=\ell ab$ for some positive integers $k,\ell$. Then, working $\pmod {c^2+1}$ we obtain $1 \equiv c(c^2-c+1) \equiv \ell ab \equiv -\ell a^2 \pmod {c^2+1},$ and so $(c^2+1) \mid (\ell a^2+1)$. Similarly, $(c^2+1) \mid (\ell b^2+1)$. Hence, we may let $\ell a^2=(c^2+1)m+c^2$ and $\ell b^2=(c^2+1)n+c^2$, for some $m,n \geq 0$. Then, $c^2(c^2-c+1)^2=(ab\ell)^2=((c^2+1)m+c^2)((c^2+1)n+c^2),$ or equivalently $(c^2+1)mn+c^2m+c^2n=c^2(c-1)^2$. Working $\pmod c^2$, we find that $c^2 \mid mn$. However, if $mn \neq 0$ this implies that $mn \geq c^2,$ and so $c^2(c-1)^2 \geq (c^2+1)mn \geq c^2(c^2+1) >c^2(c^2-2c+1)=c^2(c-1)^2,$ a contradiction. Thus, $mn=0$. WLOG assume that $m=0$, and so $n=(c-1)^2$. Note that $m=0$ implies that $a^2\ell=c^2$. However, this implies that $c(c^2-c+1)=ab\ell=ab \cdot \dfrac{c^2}{a^2},$ that is $\dfrac{b}{a}=\dfrac{c^2-c+1}{c}$. Therefore, $(c^2+1)k=b+a=\dfrac{a(c^2+1)}{c},$ which implies that $kc=a$. This, combined with $a^2\ell=c^2$ implies that $a=kc \geq c=a\sqrt{\ell} \geq a,$ so equality must hold i.e. $k=\ell=1$, and now it's easy to conclude that $a=c$ and $b=c^2-c+1$, as desired.