No such $(a,b,c)$. My solution is long and boring. Assume not. Let $x=a-2$, $y=b-2$, $z=c-2$, hence $x,y,z \ge -1$. \begin{align*} a^2+b^2+c^2+abc-2017 &= (x+2)^2 + (y+2)^2 + (z+2)^2 \\ & + (x+2)(y+2)(z+2) - 2017 \\ &= (x+y+z+4)^2 + (xyz+12) - 45^2. \end{align*}Thus the divisibility relation becomes \[ p = xyz+12 \mid \left( x+y+z+4 \right)^2 - 45^2 \]so either \begin{align*} p &= xyz+12 \mid x+y+z-41 \\ p &= xyz+12 \mid x+y+z+49 \end{align*}On the other hand, we have $(x+2)^2 + (y+2)^2 + (z+2)^2 + (x+2)(y+2)(z+2) > 2017$. Assume $x \ge y \ge z$. Claim: We have $x \ge 9$. Proof. If not, then $a,b,c \le 11$, and then $a^2+b^2+c^2+abc < 3 \cdot 11^2 + 11^3 < 2000$. $\blacksquare$ We now eliminate several edge cases to get $x,y,z \neq -1$ and a little more: Claim: We have $x,y,z > 0$. In particular $p \ge 23$, and so $x,y,z$ are all odd and not divisible by $3$. Proof. The proof involves three cases: If $x=y=z=-1$, then $p=11$, but this fails. If $x > 0$ and $y=z=-1$, then we want $p=x+12$ to divide either $x-43$ or $x+47$. We would have $x \equiv -55 \pmod p$ or $x \equiv 35 \pmod p$, but $p > 11$ contradiction. If $x,y > 0$ and $z = -1$, then $p = 12-xy > 0$. But $x \ge 9$ implies $y=1$ and $x \in \{9, 10\}$. Neither of these work. Finally, obviously $xyz \neq 0$ (else $p=12$). So $p = xyz + 12 \ge 9 \cdot 1^2 +12 = 21$, hence $p \ge 23$. Thus $\gcd(6,p) = 1$ hence $\gcd(6,xyz)=1$. $\blacksquare$ In that situation $x+y+z-41$ and $x+y+z+49$ are both even, so whichever one is divisible by $p$ is actually divisible by $2p$. Now we deduce that: \[ x+y+z+49 \ge 2p = 2xyz + 24 \implies 25 \ge xyz-x-y-z. \]We at last finish with the following cases: If $y=1$ and hence $z=1$ then $p=x+12$ should divide either $x+51$ or $x-39$. These give $39 \equiv 0 \pmod p$ or $25 \equiv 0 \pmod p$, but $p > 20$ was assumed. Otherwise, $x \ge 11$ and $y \ge 5$. \begin{align*} xyz-x-y-z &= z(xy-1) - x - y \\ &\ge xy-1 - x - y \\ &= (x-1)(y-1) - 2 > 38 \end{align*}which is a contradiction. Having exhausted all the cases we conclude no solutions exist.

Complete guess There isn't

djmathman wrote: $50 and an icecream cake that Titu wrote this

v_Enhance wrote: No such $(a,b,c)$. My solution is long and boring. Assume not. Let $x=a-2$, $y=b-2$, $z=c-2$, hence $x,y,z \ge -1$. \begin{align*} a^2+b^2+c^2+abc-2017 &= (x+2)^2 + (y+2)^2 + (z+2)^2 \\ & + (x+2)(y+2)(z+2) - 2017 \\ &= (x+y+z+4)^2 + (xyz+12) - 45^2. \end{align*}Thus the divisibility relation becomes \[ p = xyz+12 \mid \left( x+y+z+4 \right)^2 - 45^2 \]so either \begin{align*} p &= xyz+12 \mid x+y+z-41 \\ p &= xyz+12 \mid x+y+z+49 \end{align*}On the other hand, we have $(x+2)^2 + (y+2)^2 + (z+2)^2 + (x+2)(y+2)(z+2) > 2017$. Assume $x \ge y \ge z$. Claim: We have $x \ge 9$. Proof. If not, then $a,b,c \le 11$, and then $a^2+b^2+c^2+abc < 3 \cdot 11^2 + 11^3 < 2000$. $\blacksquare$ We now eliminate several edge cases to get $x,y,z \neq -1$ and a little more: Claim: We have $x,y,z > 0$. In particular $p \ge 23$, and so $x,y,z$ are all odd and not divisible by $3$. Proof. The proof involves three cases: If $x=y=z=-1$, then $p=11$, but this fails. If $x > 0$ and $y=z=-1$, then we want $p=x+12$ to divide either $x-43$ or $x+47$. We would have $x \equiv -55 \pmod p$ or $x \equiv 35 \pmod p$, but $p > 11$ contradiction. If $x,y > 0$ and $z = -1$, then $p = 12-xy > 0$. But $x \ge 9$ implies $y=1$ and $x \in \{9, 10\}$. Neither of these work. Finally, obviously $xyz \neq 0$ (else $p=12$). So $p = xyz + 12 \ge 9 \cdot 1^2 +12 = 21$, hence $p \ge 23$. Thus $\gcd(6,p) = 1$ hence $\gcd(6,xyz)=1$. $\blacksquare$ In that situation $x+y+z-41$ and $x+y+z+49$ are both even, so whichever one is divisible by $p$ is actually divisible by $2p$. Now we deduce that: \[ x+y+z+49 \ge 2p = 2xyz + 24 \implies 25 \ge xyz-x-y-z. \]We at last finish with the following cases: If $y=1$ and hence $z=1$ then $p=x+12$ should divide either $x+51$ or $x-39$. These give $39 \equiv 0 \pmod p$ or $25 \equiv 0 \pmod p$, but $p > 20$ was assumed. Otherwise, $x \ge 11$ and $y \ge 5$. \begin{align*} xyz-x-y-z &= z(xy-1) - x - y \\ &\ge xy-1 - x - y \\ &= (x-1)(y-1) - 2 > 38 \end{align*}which is a contradiction. Having exhausted all the cases we conclude no solutions exist. I basically had a huge casework bash similar to this, but my handwriting was pretty messy and cramped so I'm not sure whether any points will be taken off for that.

Whew I was so worried that I messed up some of the casework.

Darn. I solved two problems (I wasn't too sure of the wording, so I solved case where not necessarily proper divisor (3,13,31), and I wrote some stuff for when it had to be proper divisor). Terrible and boring problem.

@Evan since he is a grader Will I lose points if I missed a case or two?

I did the same exact xyz substitution as v_Enhance but how did I not think of using difference of squares

yayups wrote: @Evan since he is a grader Will I lose points if I missed a case or two? yes...

JasperL wrote: Darn. I solved two problems (I wasn't too sure of the wording, so I solved case where not necessarily proper divisor (3,13,31), and I wrote some stuff for when it had to be proper divisor). Terrible and boring problem. In my opinion, this problem was actually quite clever getting to the factorization... but then yeah the casework kind of ruined it :/

I started differently but got no solution as well. a^2+b^2+c^2+abc-2017=(a+b+c+43)(a+b+c-47)+(a-2)(b-2)(c-2)+12 So (a-2)(b-2)(c-2)+12 is a divisor of (a+b+c+43) or (a+b+c-47)

like 7- or 0+?

zephyrcrush78 wrote: JasperL wrote: Darn. I solved two problems (I wasn't too sure of the wording, so I solved case where not necessarily proper divisor (3,13,31), and I wrote some stuff for when it had to be proper divisor). Terrible and boring problem. In my opinion, this problem was actually quite clever getting to the factorization... but then yeah the casework kind of ruined it :/ Wait the factorization took 5 minutes to get to, but I didn't except it to be bashy as it is a J4

JasperL wrote: zephyrcrush78 wrote: JasperL wrote: Darn. I solved two problems (I wasn't too sure of the wording, so I solved case where not necessarily proper divisor (3,13,31), and I wrote some stuff for when it had to be proper divisor). Terrible and boring problem. In my opinion, this problem was actually quite clever getting to the factorization... but then yeah the casework kind of ruined it :/ Wait the factorization took 5 minutes to get to, but I didn't except it to be bashy as it is a J4 shoot i guess i'm stupid then the factorization took me 2 hours

I found and said it always divides when d+e+f=41 or a+b+c=47 but then said it is actually always equal when a+b+c=47 and then claimed that none work for proper. Any way this could salvage a point or 2? (Yeah I did all the mod stuff to show that a+b+c=47 satisfies divisibility) and ran out of time. Ugh I got to what jrivkin9 had but just ran out of time bc #6.

did u finish 6

Wait @zephycrush I'm 90% sure I will lose at least 1 point for my work, because I didn't really work out all the details too much. I wrote down the outline for the bounding idea (listing all the cases), and only had time to work on several of them.

yayups wrote: like 7- or 0+? depends what cases you left out, but probably still 7- if you had the factorization and bounding idea

There is no way that immigrant can know what "properly divides" mean...

trumpeter wrote:

My solution was identical except that I let $a\leq b\leq c$ to reduce case-work. Was there a reason you avoided doing this?

Is 11 a proper divisor of -2013? If so, (1, 1, 1) works.

"positive number"

Suppose in the actual test, if i just put no without any proof, what will i get?

@above 0.

Okay....

room456, your solution is completely correct! 7 points! I was thinking very similarly to you, especially getting the expressions to connect via the factoring. I was also looking for a difference of squares. Unfortunately I thought $-2*-2*-2=8$, and that's how I got a silly 2041. (This is only practice, not on the real 2017 test...) Then I went for a contradiction. Personally I find this problem very nice, because it requires a dense "hands-on" step.

Was there a way to do this one solely by creating a contradiction using mod 3? If not, modular arithmetic in general?

room456 wrote: This might be wrong idk

Nice solution; that is either a $6$ or $7$ (a $6$ is only possible since the minimum of $p$ is actually $x+4$ and achieved when we use $x-6, 3, 3$). Wait, you never asked what score you would get so ignore everything I just said since this is aops

Let $m = a^2+b^2+c^2 + abc - 2017$ and $p = (a-2)(b-2)(c-2) + 12$. Clearly, if $m$ is divisible by $p$, then $m-p$ is also divisible by $p$. $$m - p = a^ 2 + b^2 + c^2 + abc-2017 - (a-2)(b-2)(c-2)-12 = (a+b+c-2)^2 - 2025 = (a+b+c-47)(a+b+c+43).$$Let $x=a-2$, $y=b-2$, and $z=c-2$. Then, we want either $x+y+z-41$ or $x+y+z+49$ to be divisible by $xyz+12$. Assume WLOG that $x \ge y \ge z$. Claim: $x \ge 9$ Assume, by sake of contradiction, that $x < 9$. Then, $y,z < 9$ as well. Note that then $$a^2+b^2+c^2 + abc - 2017 = (x+2)^2+(y+2)^2 + (z+2)^2 -(x+2)(y+2)(z+2) - 2017 < 11^2 \cdot 3 -11^3 + 2017 < 0$$This is a contradiction because then $a^2+b^2+c^2 + abc - 2017$ is nonpositive. The problem explicitly states it to be positive. Claim: $z \ge 1$ If $z \le -2$, then $c \le 0$, which isn’t true--it’s given that $c$ is positive. If $z = -1 \implies a = 1$, then $p = -xy+12$, which must be positive. Therefore, $xy \le 12$. However, since $x \ge 9$, then $y$ must be negative. As $y \le 2$, we know that $y = -1$ is the only option. Now, if $y=-1$ and $z-1$, then $x-43$ or $x+47$ must be divisible by $x+12$, which is absurd by the Euclidean Algorithm. Therefore, $z \ne -1$. If $z=0$, then $xyz + 12=12$, which isn’t prime. Therefore, $z \ne 0$. Putting this all together, $z \ge 1$. Claim: $x+y+z-41$ can’t be divisible by $xyz+12$. We utilize proof by contradiction. Assume that $x+y+z-41$ is divisible by $xyz + 12$, then we know that $x+y+z-41 \ge xyz + 12 \implies (x-1)(y-1)(z-1) + 1 + xy + xz + yz \le -53$, which clearly isn’t true, as $(x-1)(y-1)(z-1) \ge 0$ and $xy + xz + yz \ge 0$ (recall that $z \ge 1$). Claim: $z \ne 1$ If $z=1$, then $x+y+50$ must divide $xy + 12$. If this is indeed true, then $x+y+50 \ge xy+12 \implies (x-1)(y-1) \le 39$. Given our bounds on $x$, we know that $y-1=1,2,3 \implies y =2,3,4$ because of the bounds of $x$. If $y=2,3,4$, then $xy+12$ is not prime. Therefore, $z \ne 1$ and $z \ge 2$. Claim: $x + y + z + 49$ can’t be divisible by $xyz+12$. That would imply $xyz + 12 \le x + y + z + 49 \implies 37 \ge (x-1)(y-1)(z-1) + xy + xz + zy + 1$. This clearly isn’t true given that $x \ge 9, y \ge 2, $ and $z \ge 2$. $\blacksquare$

There are no such triples $(a,b,c)$. Let $a-2=x, b-2=y, c-2=z$ and $p=xyz+12$. So $a^2+b^2+c^2+abc-2017$ $=(x+2)^2+(y+2)^2+(z+2)^2+(x+2)(y+2)(z+2)-2017$ $=x^2+4x+4+y^2+4y+4+z^2+4z+4+xyz+2xy+2yz+2xz+4x+4y+4z+8-2017$ $=x^2+y^2+z^2+8x+8y+8z+2xy+2xz+2yz+xyz+28-45^2$. Since $p|x^2+y^2+z^2+8x+8y+8z+2xy+2xz+2yz+xyz+28-45^2$, we have $p|x^2+y^2+z^2+8x+8y+8z+2xy+2xz+2yz+16-45^2$. $p|(x+y+z)^2+8(x+y+z)+16-45^2$ $p|(x+y+z+4)^2-45^2$. So $p|x+y+z-41$ or $p|x+y+z+49$. WLOG $x\ge y\ge z$. Lemma: $x,y,z>0$. Proof: Assume not. We have three cases based on how many of $x,y,z$ are negative (if one of $x,y,z$ is zero, then $p$ isn't a prime). Note that $x,y,z\ge-1$ because $a,b,c>0$. Case 1: All of $x,y,z$ are negative. So $x=y=z=-1$. We need $(x+2)^2+(y+2)^2+(z+2)^2+(x+2)(y+2)(z+2)-2017>0$, but this is not satisfied. Case 2: $x$ is positive but $y$ and $z$ are negative. So $y=z=-1$. Note that the $p=x+12$. We either have $x+12|x-43$ or $x+12|x+47$. For the former, we have $p|p-55\implies p|55 \implies p=5,11$. But this is not possible as $p>12$. The latter also doesn't hold because it implies that $p|p+35\implies p|35 \implies p=5,7$, which is not possible. Case 3: $x$ and $y$ are positive but $z$ is negative. Note that $z=-1$. So $p=12-xy$. Thus, $p<12$. If $p=11$, then $x=y=1$, but $(x+2)^2+(y+2)^2+(z+2)^2+(x+2)(y+2)(z+2)-2017>0$ is not satisfied. If $p=7$, then $x=5, y=1$, but $(x+2)^2+(y+2)^2+(z+2)^2+(x+2)(y+2)(z+2)-2017>0$ is also not satisfied. If $p=5$, then $x=7, y=1$, but $(x+2)^2+(y+2)^2+(z+2)^2+(x+2)(y+2)(z+2)-2017>0$ is not satisfied. If $p=3$, then $x=y=3$ or $x=9,y=1$, but $(x+2)^2+(y+2)^2+(z+2)^2+(x+2)(y+2)(z+2)-2017>0$ is not satisfied both times. If $p=2$, then $x=5,y=2$ or $x=10,y=1$. Both times, $(x+2)^2+(y+2)^2+(z+2)^2+(x+2)(y+2)(z+2)-2017>0$ is not satisfied. Since we have exhausted all cases, $x,y$ and $z$ are all positive and $p>12$. Lemma: $x>9$. Proof: If $x=9$, then $xyz+12$ isn't a prime. So it suffices to prove that $x<9$ is not possible. Assume that $x<9$. Then $(x+2)^2+(y+2)^2+(z+2)^2+(x+2)(y+2)(z+2)-2017<0$, a contradiction. Since $p>2$, we have $x,y,z$ are all odd. Now we will take cases how many of $x,y,z$ are $1$. Case 1: $x>1$ and $y=z=1$. This is a similar idea to $y=z=-1$ and $p=x+12$. We either have $x+12|x-39$ or $x+12|x+51$. If the former holds, we must have $p|p-51\implies p|51\implies p=13$. But this is not possible as $x>1$. If the latter holds, we must have $p|p+39\implies p=13$, but this is not possible. Case 2: $x,y>1$ and $z=1$. So $p=xy+12$. We must have $xy+12|x+y-40$ or $xy+12|x+y+50$. Note that if the former holds, we must have $x+y<40$, as $xy>x+y$. Also note that $y\ge5$ because $p$ is not prime otherwise. We need $|xy+12|=xy+12\le |x+y-40|=-x-y+40$. We also have $-x-y+40<40$, so $xy\le28$, but this is not possible as $y\ge5$ and $x>9$. If the latter holds, then we must have $xy<x+y+38$. We also have $x\ge11$, as $p$ is not a prime when $x=10$. So $(x-1)y<x+38$. Since $y\ge5$, we have $5x-5<x+38\implies 4x<43$, which is not possible. Case 3: $x,y,z>1$. Subcase 1: $xyz+12|x+y+z-41$. If both $xyz+12$ and $x+y+z-41$ are positive, this won't hold as $xyz>x+y+z$. So $x+y+z-41\le0$. We need $|xyz+12|=xyz+12<|x+y+z-41|=-x-y-z+41$. Thus, $xyz+12<41$, which is not possible. Subcase 2: $xyz+12|x+y+z+49$. So $xyz<x+y+z+37$. Thus, $x(yz-1)<y+z+37$. Since $x\ge11$, we have $11yz-11<y+z+37$, so $11yz<y+z+48$. Therefore, $z(11y-1)<y+48$. Since $z\ge5$, we have $55y-5<y+48\implies 54y<53$, a contradiction. So the answer is $\boxed{\text{no}}$.

The answer is no. Substitute $x=a-2,y=b-2,z=c-2$. This means that $x,y,z\geq -1$. Then $$a^2+b^2+c^2+abc-2017=(x+y+z-41)(x+y+z+49)+xyz+12.$$It is given in the problem that this is positive. Now, suppose for the sake of contradiction that $xyz+12$ is a prime. Clearly $x,y,z\neq 0$. Then we have $$\frac{(x+y+z-41)(x+y+z-49)}{xyz+12}$$is an integer greater than or equal to $1$. This also implies that $x+y+z > 41$. Since $xyz+12$ is prime, we must have $$xyz+12\mid x+y+z-41\text{ or } xyz+12\mid x+y+z+49.$$Additionally, $x, y, z$ must be odd, so that $xyz+12$ is odd while $x+y+z-41,x+y+z+49$ are even. So, if $$xyz+12\mid x+y+z-41\text{ or }xyz+12\mid x+y+z+49,$$we must have $$2(xyz+12)\leq x+y+z-41\text{ or }2(xyz+12)\leq x+y+z+49.$$Now suppose WLOG that $x=-1$ and $y,z>0$. Then we must have $yz\leq 10$, impossible since $x+y+z>41$. Again, suppose that $x,y=-1$ and $z>0$. Then we must have $$2(z+12)\leq z-43\text{ or }2(z+12)\leq z+47,$$and since in this case we must have $z>43$, this is also impossible. Then the final case is when $x,y,z$ are positive odd numbers. Note that if $xyz>x+y+z$ for positive integers $x,y,z$, then $abc>a+b+c$ for positive integers $a,b,c$ where $a>x,b>y,c>z$. Then we only need to prove the case where $x+y+z=43$, since $x+y+z$ is odd. Then one of $$2(xyz+12)\leq 2\text{ and/or }2(xyz+12)\leq 92$$is true, implying that $xyz\leq -11$ or $xyz\leq 34$. But if $x+y+z=43$, then $xyz$ is minimized when $x=1,y=1,z=41$, so that $xyz\geq 41$. This is a contradiction, so we are done. bash

resolv No such triples exist. Suppose there did exist such a triple $(a,b,c)$. Let $x = a-2, y = b-2, z = c-2$ and $p = xyz + 12$ ($p$ is prime). We have\begin{align*} S = a^2 + b^2 + c^2 + abc - 2017 = (x+2)^2 + (y+2)^2 + (z+2)^2 + (x+2)(y+2)(z+2) - 2017 \\ = xyz + x^2 + y^2 + z^2 + 2(xy + yz + zx) + 8(x + y + z) - 1997 \\ =xyz (x+y+z)^2 + 8(x+y+z) - 1997 \\ \end{align*}Since $p\mid S$, we have $p\mid S - (xyz + 12)$, which equals\[(x+y+z)^2 + 8(x+y+z) - 2009 = (x+y+z + 49)(x + y + z - 41)\]Since $S$ is a positive multiple of $xyz + 12$ greater than $xyz + 12$, we must have $(x+y+z + 49)(x + y + z - 41)$ positive, so $x + y + z > 41$. Claim: $x,y,z$ are all positive. Proof: Suppose this wasn't the case. Then $z = -1$ must be true (since none of the variables can be zero). If $y$ was positive, then $p = 12 - xy < 12$. However, we have $x + y + z > 41$, so $x + y > 42$ meaning that at least one of $x,y$ is over $21$, so $xy > 12$, contradiction because it means $p$ is negative. Therefore, $y = -1$ also. Then, $x + 12 \mid (x + 47)(x - 43)$. Since $x + 12$ is a prime, either $x + 12 \mid x + 47$ or $x + 12 \mid x - 43$, so $x + 12$ divides $35$ or $55$, both are bad since $x + 12$ is a prime at least $12$ (since $x, y, z = -1$ is obviously bad). $\square$ If $p\mid x + y + z - 41$, then since $x + y + z - 41 > 0$, we have $x + y + z > p + 41 = xyz + 53$. However, we see that $xyz + 53 > x + y + z$ for all positive integers $x,y,z$ (since it's true for $(1,1,1)$ and stays true when you increase any variable by $1$). Therefore, $p$ cannot divide $x + y + z - 41$. Therefore, $p$ must divide $x + y + z + 49$, so\[ xyz + 12 \mid x + y + z + 49\]We have that $xyz > x + y + z - 3$ for all positive integers $x,y,z$ since it's true for $(1,1,1)$ and stays true when you increase any variable by $1$. Notice that $4(xyz + 12) > 4(x + y + z - 3) = 4(x+y+z) - 12 > x + y + z + 49$ (since $x + y + z > 41$). Hence $\frac{x + y + z + 49}{xyz + 12}$ is even and less than $4$, so it must be $2$, so\[ 2xyz = x + y + z + 25\]Clearly $2xyz - (x + y + z)> 2(x + y + z - 3 ) - (x+y+z) = x + y + z - 6 > 25$, contradiction.