the ordered triplets (a,b,c) are :- i) (a,a-1,a) ii) (a,1,a(a-1)) this was probably the easiest of all the postal problems.

Again, please use LaTeX, and don't just post answers but rather complete proofs.

Rushil wrote: Fins all ordered triples $(a,b,c)$ of positive integers such that $abc + ab+ c = a^3$ Okey so let me try the complete sol though it seems bad way , may be someone else like venkata know nicer and better solutions : We show all the ordered triplets $(a,b,c)$ are : i) $(a,a-1,a)$ ii) $\Big(a,1,a(a-1)\Big)$ We have $a(a^2-b)=c(1+ab)$ cause $\gcd(a,ab+1)=1$ then $a\mid c$ we put $c=ak$ where $k\in Z_+$ and $a^2-b=(ab+1)k$ so $a^3+1=bka^2+ab+ka+1=(ka+1)(ab+1)$. Now use Four Number Theorem we have $x,y,z,t$ which $(x,z)=1$ and $a+1=xy,a^2-a+1=zt,ka+1=zy,ba+1=xt$. Because $k,b< a$ so $zy\leq a(a-1)+1=zt$ so $y\leq t$. Other hand $xt\leq (a-1)a+1=zt$ so $x\leq z$. If $z>x$ and $t>y$ then $k,b<a-1$ so $ka+1,ba+1\leq (a-1)^2=(xy-2)^2<(xy)^2$ so $xyzt<(xy)^2$ then $zt<xy$, contract!. So $z=x$ or $t=y$ 1) IF $z=x$ then $b=a-1$ and $ka+1=zy=xy=a+1$ then $k=1$ so $c=a$ we have got $(a,b,c)=(a,a-1,a)$. 2) IF $t=y$ then $k=a-1$ so $c=a(a-1)$ and then $ba+1=xt=xy=a+1$ so $b=1$. We have got $(a,b,c)=\Big(a,1,a(a-1)\Big)$.

Maybe I do something wrong. We have $a|c$. From equation we have $a^2>c\geq a>b$ then let $c=ad$ then we have $d<a$ then the equation comes to $abd+b+d=a^2$ then we have $a|b+d$ If $b+d>a$ contradiction! Then we have $b+d=a$ the equation comes to $bd+1=a$. From this the solutions is $(a,b,c)=(xy+1,x,(xy+1)y)$

The equation is equivalent to $a^3+1=abc+ab+c+1=(ab+1)(c+1)$. Then \[ \\ ab+1 \mid a^3+1 \\ \\ ab+1 \mid a^3-ab \\ \\ ab+1 \mid a^2-b \\ \\ ab+1 \mid a^2b-b^2=(ab)a-b^2 \\ \\ ab+1 \mid b^2+a \\ \\ab+1 \mid b^3+ab \\ \\ab+1 \mid b^3-1. \\ \] Suppose $b \geq a^2$. Then, in the original equation, $ab+1 \geq a^3+1=RHS$, so we need $c+1 \leq 1, c \leq 0$ for equality, absurd because $c$ is positive. Then $b<a^2$, so $a^2-b>0$. Since $ab+1 \mid a^2-b$, we have $ab+1 \leq a^2-b$. If $b \geq a$, we have $a^2+1 \leq LHS \leq a^2-b$, absurd. So $b<a$. Since $ab+1 \mid b^2+a$, we have $ab+1 \leq b^2+a$. Since $b<a$, let $a=b+r$ with $r$ positive. Then $b^2+br+1 \leq b^2+b+r$, so $(b-1)(r-1)=br-b-r+1 \leq 0$. Then one of $b, r$ is at most $1$. If $b \leq 1$, since $b>0$ we have $b=1$, and in the original equation $(a+1)(c+1)=(a^3+1)$, so $c=a^2-a$ and this is a solution. If $r \leq 1$, we have $r=1$, $a-1=b$, so $ab+1=a^2-a+1$ and $c+1=a+1, c=a$. Solutions are $(k, 1, k^2-k)$ and $(k, k-1, k)$.