Any idea?

$\frac{(a-b)(b-c)(c-a)}{2}+2=2016^n\Leftrightarrow (a-b)(b-c)(c-a)=2\cdot 2016^n-4$ if $n=0$ $(a-b)(b-c)(c-a)=-2$ $a-b\ge b-c \ge c-a$ we meet 2 cases 1. $a-b=b-c=1 , c-a=-2\implies (a,b,c)=(k+2,k+1,k)$ 2. $a-b=2,b-c=1,c-a=-1$ but $-1\not -3$ so no solution here for $n\ge 1$ we have $4|2\cdot 2016^n - 4$ if $a,b,c$ consist of 2 even numbers + 1 odd number of 1 even number + 2 odd numbers we cannot have $4|(a-b)(b-c)(c-a)$ so $a,b,c$ all odd or all even this makes $8|(a-b)(b-c)(c-a)$ but $8\nmid 2\cdot 2016^n -4$ so no such triplet (or quadruplet) exist here

Reynan wrote: $\frac{(a-b)(b-c)(c-a)}{2}+2=2016^n\Leftrightarrow (a-b)(b-c)(c-a)=2\cdot 2016^n-4$ we cannot have $n=0$ since $LHS\ge 2$ so we have $n\ge 1$ because $n\ge 1$ we have $4|2\cdot 2016^n - 4$ if $a,b,c$ consist of 2 even numbers + 1 odd number of 1 even number + 2 odd numbers we cannot have $4|(a-b)(b-c)(c-a)$ so $a,b,c$ all odd or all even this makes $8|(a-b)(b-c)(c-a)$ but $8\nmid 2\cdot 2016^n -4$ so no such triplet (or quadruplet) exist CMIIW $\text{Are you sure?}$

The solutions is $(a,b,c)=(k+2,k+1,k)$ where $k\in\mathbb{Z}.$

thank you Ferid

I think the only solution $n=0$ My solution: If $n\geq 1$ then $RHS\equiv 0 (\mod9)$ But $LHS\equiv 2 (\mod 9)$ Then $n=1$ $(a-b)(b-c)(c-a)+4=2\cdot 2016^n$

$x=a-b...$ then $x+y+z=0$ and $xyz=2\cdot 2016^n-4$ we have $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)=0$ thus $x^3+y^3+z^3=6 \cdot 2016^n-12...(*) $ so for $n \ge 1$ we get $x^3+y^3+z^3 \equiv 2 (mod 7)$ wlog let $7|x$ and $y^3,z^3 \equiv 1 (mod 7)$ but we have $7|0=x+y+z$ thus $7|y+z$ and $y^3,z^3 \equiv 1(mod 7)$ if $a^3 \equiv 1 (mod 7)$ we have $a \equiv 1,2,4 (mod 7)$ so $y,z \equiv 1,2,4 (mod 7)$ but with this we can't have $7|y+z$ so $n=0$ now it is rather easy

Nice solution SidVicious! This problem was proposed by Greece, Silouanos Brazitikos (me)

Here is my idea: we can add an integer to all of a and b and c and then the equation would be the same. so we can make one of a and b and c to be zero. so by putting c=0 then the equation would be much more easier. and it would be a quadratic equation by a. so the delta should be a perfect square ...

I think that this problem was the easiest on the competition, but I couldn't solve it. I made a stupid mistake when was using mod 9

MilosMilicev wrote: I think that this problem was the easiest on the competition, but I couldn't solve it. I made a stupid mistake when was using mod 9 Were you contestant4 from Bulgaria?

No, SRB 5.

Aa ok!!!

(WRONG solution, thanks to Murad.Aghazade) if at least two numbers of a,b,c are equal, then we get a contradiction . so a,b,c are different now WLOG assume that $a>b>c$ so $a-b>0 , b-c>0 , c-a<0$ so $(a-b)(b-c)(c-a)<0$ let $(a-b)(b-c)(c-a)=-k$ , $k$ positive integer $-k/2 +2 >0$ so $k<4 $ , imply that $(a-b)(b-c)(c-a)= -3,-2,-1$ since $(a-b)(b-c)(c-a)$ is even , we cannot have $(a-b)(b-c)(c-a)= -3, -1$ so $(a-b)(b-c)(c-a)= -2 ; n=0 $ from $a-b>0 , b-c>0 , c-a<0 $so the answer is $(m+2,m+1,m)$ (m=integer) and the permutations(because we assume that a>b>c)

reveryu wrote: if at least two numbers of a,b,c are equal, then we get a contradiction . so a,b,c are different now WLOG assume that $a>b>c$ so $a-b>0 , b-c>0 , c-a<0$ so $(a-b)(b-c)(a-c)<0$ let $(a-b)(b-c)(a-c)=-k$ , $k$ positive integer $-k/2 +2 >0$ so $k<4 $ , imply that $(a-b)(b-c)(a-c)= -3,-2,-1$ since $(a-b)(b-c)(a-c)$ is even , we cannot have $(a-b)(b-c)(a-c)= -3, -1$ so $(a-b)(b-c)(a-c)= -2 ; n=0 $ from $a-b>0 , b-c>0 , c-a<0 $so the answer is $(m+2,m+1,m)$ (m=integer) and the permutations(because we assume that a>b>c) $(a-b)(b-c)(c-a)$ isn't cyclic. Because if you let $a$----->>>$b$ So you will not get the same thing. So you can't say WLOG...

silouan wrote: Nice solution SidVicious! This problem was proposed by Greece, Silouanos Brazitikos (me) Congrats on proposing such great problem (but is easy for p3) Did you send this as an easy problem or what? İn Senior Balkan your easy problem were selected p3.

Murad.Aghazade wrote: silouan wrote: Nice solution SidVicious! This problem was proposed by Greece, Silouanos Brazitikos (me) Congrats on proposing such great problem (but is easy for p3) Did you send this as an easy problem or what? İn Senior Balkan your easy problem were selected p3. You said that his problem is easy for 3 times. We got it.

Pure_IQ wrote: Murad.Aghazade wrote: silouan wrote: Nice solution SidVicious! This problem was proposed by Greece, Silouanos Brazitikos (me) Congrats on proposing such great problem (but is easy for p3) Did you send this as an easy problem or what? İn Senior Balkan your easy problem were selected p3. You said that his problem is easy for 3 times. We got it. It is interesting that for the second time easy problem of Mr.Silouanos is selected as p3 of BMO and JBMO.

They are secret till the next JBMO.

Reynan wrote: $\frac{(a-b)(b-c)(c-a)}{2}+2=2016^n\Leftrightarrow (a-b)(b-c)(c-a)=2\cdot 2016^n-4$ if $n=0$ $(a-b)(b-c)(c-a)=-2$ $a-b\ge b-c \ge c-a$ we meet 2 cases 1. $a-b=b-c=1 , c-a=-2\implies (a,b,c)=(k+2,k+1,k)$ 2. $a-b=2,b-c=1,c-a=-1$ but $-1\not -3$ so no solution here for $n\ge 1$ we have $4|2\cdot 2016^n - 4$ if $a,b,c$ consist of 2 even numbers + 1 odd number of 1 even number + 2 odd numbers we cannot have $4|(a-b)(b-c)(c-a)$ so $a,b,c$ all odd or all even this makes $8|(a-b)(b-c)(c-a)$ but $8\nmid 2\cdot 2016^n -4$ so no such triplet (or quadruplet) exist here Is this solution right?

@phymaths yes,Reynan had edited it. @Murad.Aghazade Thank you very much.

This one is quite easy, you just split it in n=0 and n>0, and you get solution for the first case and you simply can prove that the second one does have the solution by (mod 7). The second case (first case is just a bit too easy and even a beginner like me was able to find it in less than a minute): WLOG b > a > c; b - a = x a - c = y b - c = x + y Now we have 2016^(m) = (-x)(-y)(x+y)/2 + 2 2016^(m) = xy(x+y)/2 + 2 Because of the 2016 === 0 (mod 7) it must be xy(x+y) === 3 (mod 7), and this is impossible, so we have only the solutions from the first case. This was really easy and definitely should not find place on a shortlist, although solving this was quite nice because of the use of mod 7.

Inmo problem $3$ copied from this 2017 $Inmo=Indian National Mathematics Olympiad $

Ferid.---. wrote: I think the only solution $n=0$ My solution: If $n\geq 1$ then $RHS\equiv 0 (\mod9)$ But $LHS\equiv 2 (\mod 9)$ Then $n=1$ $(a-b)(b-c)(c-a)+4=2\cdot 2016^n$ It is true that LHS can never be $0 (\mod9)$ though you have to explain/prove why. However, there are many counterexamples which can show that LHS is not always $2 (\mod 9)$, provided you only consider $3^2$ (and not the other unique prime factors of 2016). One counterexample is the following: $LHS = \frac{(-4)(-1)(5)}{2} + 2= 12 \equiv 3 (\mod9) $

phymaths wrote: Reynan wrote: $\frac{(a-b)(b-c)(c-a)}{2}+2=2016^n\Leftrightarrow (a-b)(b-c)(c-a)=2\cdot 2016^n-4$ if $n=0$ $(a-b)(b-c)(c-a)=-2$ $a-b\ge b-c \ge c-a$ we meet 2 cases 1. $a-b=b-c=1 , c-a=-2\implies (a,b,c)=(k+2,k+1,k)$ 2. $a-b=2,b-c=1,c-a=-1$ but $-1\not -3$ so no solution here for $n\ge 1$ we have $4|2\cdot 2016^n - 4$ if $a,b,c$ consist of 2 even numbers + 1 odd number of 1 even number + 2 odd numbers we cannot have $4|(a-b)(b-c)(c-a)$ so $a,b,c$ all odd or all even this makes $8|(a-b)(b-c)(c-a)$ but $8\nmid 2\cdot 2016^n -4$ so no such triplet (or quadruplet) exist here Is this solution right? In fact, no. Though it is true that $n>0$, $4|(a-b)(b-c)(c-a)$ may hold with 1 even numer + 2 odd numbers. For example, $(a, b, c) = (4, 0, 1)$. Thus, we can say this solution is incorrect.

Sorry for bumping this, but can we assume WLOG a>=b>=c in this problem?If not,can you explain in full detail?I am a novice.

bump?...

MathematicalPhysicist wrote: Sorry for bumping this, but can we assume WLOG a>=b>=c in this problem?If not,can you explain in full detail?I am a novice. no you can't if $a >= b >=c$ then $(a-b)(b-c)(c-a)$ is non-positive if $a >= c >=b$ then $(a-b)(b-c)(c-a)$ is non-negative

If we take mod9 in n>=1 everything is clear. After we find n=0 we can take cases because it is a product so it can have only +2, +1, -1, -2. WLOG we assume that a-b>=b-c>=c-a. The conclusion follows...

Apply am gm and we get n value is only 0 and the rest follows

for n = 0 we get (a,b,c) = (d+2,d+1,d) (we can manually solve (a-b)(b-c)(c-a) = -2 now suppose n>=1 (a-b)(b-c)(c-a) = 2 * (2016^n - 2) a^{2}*(c-b) + c^{2}(b-a) + b^{2}(a-c) = 2 * (2016^n -2) RHS is congruent to 2 mod 3, hence a,b and c should all leave different remainders modulo 3(otherwise u get 3|LHS) now WLOG suppose a = 3k +1 b = 3m + 2 c = 3x RHS is congruent to 5 mod 9 LHS = (3k - 3m - 1)(3m - 3x + 2)(3x - 3k - 1) = (-27k^{2}m+27k^{2}x+27km^{2}-27kx^{2}-27m^{2}x+27mx^{2}-18k^{2}+18km+18kx+9m^{2}-36mx+9x^{2}+9m-9x+2) which is clearly congruent to 2 mod 9, contradiction! which means no solutions for n>=1 hence the only solutions are (a,b,c) = (d+2,d+1,d) which is achieved when n=0

We work with $(a-b)(b-c)(c-a) = 2 \cdot (2016^k - 2)$, where $k$ is a non-negative integer. Note that $2016 = 2^5 \cdot 3^2 \cdot 7$. Write for convenience $x = a-b$, $y=b-c$, so $xy(x+y) = 4 - 2 \cdot 2016^k$. If $k\geq 1$, then $xy(x+y) \equiv 4 \pmod 7$, which is impossible by direct computation. (Alternatively, one can directly show that $xy(x+y) \equiv 4 \pmod 9$ never holds, a slightly quicker way is firstly to work mod 3 and obtain $x \equiv y \equiv 1 \pmod 3$.) So $k=0$ remains, i.e. $xy(x+y) = 2$. If $x=2$, then $y^2+2y-1 = 0$ with no solutions; if $x=1$, then $y=1$ or $y=-2$; if $x=-1,$ then $y^2 - y + 2 = 0$ with no solutions and if $x=-2$, then $y=-1$. Hence $(a,b,c) = (k+2,k+1,k)$ or $(k+1,k,k+2)$ or $(k,k+2,k+1)$, where $k$ is an arbitrary integer.