Cyclic powers of 2! IMO 2015 problem#2

If $a=1$ ,then $c-b$ and $b-c$ both are powers of 2.It is impossible.

So $a≥2$

Similarly $b≥2,c≥2$

Here are two cases discussed.

Case 1 There are two equal numbers at least in them.

Might as well assume $a=b$

We have $ac-b=a(c-1)$ is powers of 2.

$∴ a,(c-1)$ both are powers of 2.

Let $a=2^p,c=1+2^q(p≥1,q≥0)$ ,then $ab-c=2^{2p}-2^q-1$ is power of 2.

If $q＞0$ , $2^{2p}-2^q-1$ is odd,

$∴2^{2p}-2^q-1=1$

$∴2^q≡2(mod 4)$

$∴q=1,p=1,a=2,b=2,c=3.$

If $q=0$ , $2^{2p}-2$ is power of 2,only $p=1$ satisfied this case.

So, $a=2,b=2,c=2.$

We are easy to verify $(2,2,2),(2,2,3)$ both are satisfied.

Case 2 a,b,c are different from each other

Might as well assume $2≤a＜b＜c$

and $bc-a=2^α,ac-b=2^β,ab-c=2^γ,(α＞β＞γ≥0)$

1° $a=2$

We prove $γ=0$ at first.

If $γ＞0$ ,from $ab-c=2^γ$ ,we have that $c$ is even.From $ac-b=2^β$ ,we have that $b$ also is even.

Therefore, $bc-a≡2(mod 4)$ ,but $2^α≡0(mod 4)$ .Paradoxical!

$∴γ=0$

$∴c=2b-1$

Since $ac-b=3b-2=2^β≡1(mod 3)$ , $β$ is even.

If $β=2$ ,then $b=2=a$ .False.

If $β=4$ ,then $b=6,c=11$ .

We are easy to verify $(2,6,11)$ is satisfied.

If $β≥6$ ,then $b=\frac{1}{3}$ $(2^β+2)$ .Substituted it in $bc-a$ and we have

$9 \times 2^α=9(bc-a)=9b(2b-1)-18=(3b-2)(6b-1)-16=2^β(2^{β+1}+5)-16$ ①

$∵α＞β≥6$

$∴2^7|9 \times 2^α$ ,but Eq.① only divisible by $2^4$ and not divisible by $2^5$ ,Paradoxical!

2° $a≥3$

We have $(a+b)(c-1)=2^α+2^β,(b-a)(c+1)=2^α-2^β$ .

∵one of $c+1$ and $c-1$ is not a multiple of 4

∴ $2^{β-1}|(a+b)$ or $2^{β-1}|(b-a)$

$∵2^β=ac-b≥3c-b＞2c$

$∴b＜c＜2^{β-1}$

$∴0＜b-a＜2^{β-1}$

$∴2^{β-1}|(b-a)$ is false.

∴ $2^β|(a+b)$

And $a+b＜2b＜2^β$ ,so $a+b=2^{β-1}$ .

$∴ac-b=2^β=2(a+b)$

$∴a(c-2)=3b$

If $a≥4$ ,then $b≥5$

$∴a(c-2)≥4(c-2)≥4(b-1)＞3b$ .Paradoxical!

$∴a=3,c-2=b$

$∴b=2^β-3,c=2^β-1$

$∴2^γ=ab-c=3(2^β-3)-(2^β-1)=2^β-8$

$∴β=4,b=5,c=7$

We are easy to verify $(3,5,7)$ is satisfied.

In summary,there are $\boxed{4}$ three-number sets satisfied.They are $(2,2,2)$ , $(2,2,3)$ , $(2,6,11)$ , $(3,5,7)$

I don't know whether IMO would encourage this or not, but i sat down and tried to find out all solutions. It turned out to be only 4. (2,2,2) (2,3,2) (3,5,7) (2,6,11).

Proof of (2,2,2). Let a=b=c

Therefore, a(a-1)=0(mod 2)

There's only 1 solution possible for this argument Hence one solution is (2,2,2)

I'll surely post solution for the rest three when I get them!

Please do post a proper solution.