Maybe hard than IMO 2015 problem 2 Find all postive integers $(a,b,c)$ ,  such that$a^2b-c,b^2c-a,c^2a-b$ are all powers of  2
someone can take a example such this condition
 A: Note: This is only a proof sketch. I omitted some details which are not really technical and you can verify it by your own. If there are some mistakes please tell me.

Let us assume that $$a^2b-c=2^p,\ b^2c-a=2^q,\ c^2a-b=2^r$$ and within them $r$ is the smallest. To simplify the discussion I further assume here 'powers of $2$' means that $p,q,r>0$ (or we'll have something like $c=a^2b-1$ and these cases needs more effort to probe).
Denote $x=bc,\ y=ac,\ z=ab$, and we can write
$$z^2=2^p\cdot b+x,\ x^2=2^q\cdot c+y,\ y^2=2^r\cdot b+z,$$
so we have
$$z^2\equiv x,\ x^2\equiv y,\ y^2\equiv z\ (\mathrm{mod}\ 2^r),$$
and therefore using a simple induction one can deduce that $x\equiv y\equiv z\equiv 0\textrm{ or }1\ (\textrm{mod }2^r)$. Within both cases modulo $2^r$ on the original equations will lead to $a\equiv b\equiv c\ (\textrm{mod }2^r)$. They are the quadratic root of $x$ modulo $2^r$, which actually has only five possibilities:
$$a\equiv b\equiv c\equiv 0\textrm{ or }\pm1\textrm{ or }\pm1+2^{r-1}\ (\textrm{mod }2^r).$$
Before we discuss these posibilities, there is a useful observation that $x,y,z$ cannot be $0$ nor $1$. So it follows that $ac=y\geq 2^r=c^2a-b$, which means $b\geq ac(c-1)$ is much larger than $a$ and $c$ (assuming, of course, that $a,b,c>1$). It actually implies that $p,q>r$.


*

*If it is $0$, notice that we can deduce from the original equations that
$$(ab^3-1)c=2^p+2^q\cdot ab$$
which means that the smaller one of $2^p$ and $2^q$ is a factor of $c$. But either $c\geq 2^p$ or $c\geq 2^q$ will lead to contradiction with the above observation that $b$ is large.

*If it is $\pm1$, notice that $z=y^2-2^r\cdot b\equiv 1+2^r\ (\textrm{mod }2^{r+1})$. On the other hand $x,y\equiv 1\ (\textrm{mod }2^{r+1})$ since $p,q>r$. Let us denote $a=a_1\cdot 2^r\pm 1$, and $b_1,c_1$ similarly.
Observe that $ab\equiv 1\ (\textrm{mod }2^{r+1})$ if and only if $a_1+b_1$ is even. However, the above discussion shows that $2$ numbers amongst $a_1+b_1,\ a_1+c_1,\ b_1+c_1$ is even, which is a contradiction.

*If it is $\pm1+2^{r-1}$, which are the cases we only consider when $r\geq 3$. We define $a_1,b_1,c_1$ similarly but this time there is no direct contradiction since $ab\equiv 1\ (\textrm{mod }2^{r+1})$ if and only if $a_1+b_1$ is odd. Instead, we will be informed that


*

*$b_1$ is even and $a_1,c_1$ is odd;


or


*

*$b_1$ is odd and $a_1,c_1$ is even.


Now cosider the equation $c^2a-b=2^r$. Modulo $2^{r+1}$ we will have
$$(2^r+2^{r-1}\pm 1)^3-(2^{r-1}\pm 1)\equiv 2^r\ (\textrm{mod }2^{r+1})$$
or
$$(2^{r-1}\pm 1)^3-(2^r+2^{r-1}\pm 1)\equiv 2^r\ (\textrm{mod }2^{r+1})$$
which cannot hold when $r\geq 3$.

Finally it remains the case when some of $a,b,c$ is $1$, and from the above discussion we see this could only happens when $p=r$ or $q=r$. You can exhaust the possibilities and when I do that, I find no solutions. So my conclusion is that there is no solution, if we read 'powers of 2' excluding $2^0=1$.
A: Another sketch of a proof. Key thing is that either $a,b,c$ are all odd or all even$
Assume they are all even, $a = 2a_1, b = 2b_1, c = 2c_1$
$4a_1^2 b_1 - c_1 = 2^{m-1} $, $4b_1^2c_1 - b_1 = 2^{n-1} $, $4c_1^2a_1 - b_1 = 2^{p-1} $,
Again all $a_1, b_1, c_1$  must be even and we will descend until each of the three equations are as in
case 1: $a_1^2b_1 - c_1 = 1$ or 
case 2: $a_1^2b_1 - c_1 = 2^k$ (latter looks like the original equation).
If all the equations are as in case 1, then no solution. If they are all in case 2, we are back to the original form and continue. 
Now there is a possibility of mix and match. Suppose first equation is as in case 1 and second two are as in case 2. We have:
$a_1^2b_1 - c_1 = 1,\ \ b_1^2c_1 - a_1 = 2^k, \ \ c_1^2a_1 - b_1 = 2^l$
If $c_1$ is odd, then one of $a_1, b_1$ has to be even. 
Assume $a_1$ is even, $b_1$ odd. Then $c_1^2a_1 - b_1 = 2^l$ is impossible
Assume $a_1$ is even, $b_1$ even. Continuing we will end up with either an impossible case or all of $a,b,c$ are odd. 
Assume $a,b,c$ all odd:
$(2a_1+1)^2 (2b_1+1) - (2c_1+1) = 2^m $,
$(2b_1+1)^2 (2c_1+1) - (2a_1+1) = 2^n $,
$(2c_1+1)^2 (2a_1+1) - (2b_1+1) = 2^p$,
Simplifying we get for the first equation: $4a_1^2b_1+4a_1b_1 + 2a_1^2 + 2a_1 +b_1 - c_1 = 2^{m-1}$
Again all of $a_1, b_1, c_1$ are all odd or all even until they (each of $a_1, b_1, c_1$) end up at one or RHS becomes $2^0 = 1$. There doesn't seem to be a solution in this case. 
