Our Equation \[ (x - a)(x - b)(x - c)[(x - a)^i(x - b)^j(x - c)^k - 1]\] is always $ 0 \pmod p$. First we may assume that $ 0 \le i,j,k < p-1$ because if $ i,j,k$ any is greater than or equal to $ p-1$ we may replace it with $ i-p+1, j-p+1, k-p+1$, still preserving the value of $ i+j+k \pmod {p-1}$ Since these are each less than $ p-1$, $ i+j+k < 3(p-1)$ If $ i+j+k = 0$ then we are done. Assume for sake of contradiction they are not. If $ i+j+k = 2(p-1)$ we substitute $ p-1-i, p-1-j, p-1-k$ for $ i,j,k$ and since this is the inverse of the previous expression $ \pmod p$, which is always $ 1 \pmod p$ for all $ x \ne a,b,c$, the whole polynomial is still $ 0 \pmod p$ for all integers. So we can assume $ i+j+k = p-1$. Also since the equation is cyclic assume $ i \ge j, i \ge k$ and therefore $ i \ge {(p-1) \over 3}$ Also assume $ p>7$, we have $ (x - a)(x - b)(x - c)[(x - a)^i(x - b)^j(x - c)^k - 1]$ $ = (x - a)(x - b)(x - c)[(x - a)^i(x - b)^j(x - c)^k - (x-a)^{p-1}]$ $ = (x - a)(x - b)(x - c)[(x-b)^j(x-c)^k - (x-a)^{j+k}]$ (We can divide out by all the repeated roots because the polynomial remains $ 0 \pmod p$ for all integers.) Now this polynomial has degree $ \le {2p+7 \over 3}$ which for all $ p>7$ is less than $ p$ thus this polynomial (which is clearly not the 0 polynomial since $ a,b,c$ are distinct) has less than $ p$ roots $ \pmod p$ thus it is not $ 0 \pmod p$ for all integers $ x$. For the $ p=7$ case the above argument holds for all cases except $ i=j=k=2$. then we have \[ (x-a)(x-b)(x-c)[(x-b)^2(x-c)^2-(x-a)^4]\] Since the $ x^4$ term within the brackets vanishes the polynomial is of degree 6, and is clearly not the zero polynomial $ \pmod 7$. Contradiction

Wonderful solution! Thank you. I was trying to use primitive roots which failed.

in the field $ Z_{p} $ is still true that if $ f $ has a double root $ r $ then $ f'(r)=0 $?

anonymouslonely wrote: in the field $ Z_{p} $ is still true that if $ f $ has a double root $ r $ then $ f'(r)=0 $? Yes, this is true. It follows easily from the fact that the product-rule for derivatives works the same in $Z_p[x]$ as it works in $\mathbb{R}[x]$.

Suppose $f(x)=(x-a)(x-b)(x-c)[(x-a)^i(x-b)^j(x-c)^k-1]$.Now putting $x=1,2,.....,p$ and calling $(t-a)\equiv a_t(mod p)$ and $(t-b)\equiv b_t (mod p)$ also $(t-c)\equiv c_t (mod p)$ we obtain $p|(a_t)(b_t)(c_t)[(a_t)^i(b_t)^j(c_t)^k-1]$ where $\{a_t\}=\{b_t\}=\{c_t\}=\{1,2,3,.....,p\}$. Now suppose $\{1,2,3,....,p\}=\{g^{a_1},g^{a_2},........,g^{a_p}\} (mod p)$. Also call $a_t \equiv g^{f_{1}(t)}(mod p )$ also $b_t\equiv g^{f_{2}(t)}(mod p )$ and $c_t\equiv g^{f_{3}(t)}(mod p )$ for all $1\leq t\leq p$.Now certainly we're getting $p-1|if_{1}(t)+jf_{2}(t)+kf_{3}(t)$ for all such $t$.Now $WOLG$ assume $i\geq j\geq k$. Also note as we've $i+j\equiv -k (mod (p-1))$ so we get $p-1|mf_{2}(t)+nf_{3}(t)$ for all such $t$ with $m=i-j,n=i-k$. Now consider a mapping where if $f_{2}(t)=h$ then $f_{3}(t)=X(h)$ for all $1\leq h\leq p$. Thus now for all $h$ we've $p-1|hx+yX(h)$. Now so we've if $p-1$ doesn't divide $y$ then $p-1|(k-1)X(k)-kX(k-1)$ also so, $p-1|X_{p-1}-(p-1)X(1)$.Now basically just summing we obtain $\frac{p(p-1)}{2}-Xp\equiv \frac{p(p-1)}{2} X(1) (mod p)$ and so $p|Xp$.Now note $p-1|pX_{p-1}-p(p-1)$ and obviously so $X_{p-1}=p-1$. Now similarly going on we get $X_{i}=i$ for all $1\leq i\leq p$ if our $y$ is not divisible by $p$.Now as $p>5$ so we can chose a odd prime $q<p$ .Now so we've $p-1|q(x+y)$ but that implies $p-1|x+y$ since $q$ is odd and $gcd(p-1,q)=1$. And hence $p-1|i+j-2k$ also we've $p-1|i+j+k$ and so finally we get $p-1$ divides all of $i,j,k$.

Solution. By Fermat's Little Theorem (from here one, we will call it FLT) we see that only considering the residue classes of $i,j,k$ modulo $p-1$ doesn't affect the congruence. Thus, we may assume that $i,j,k$ are in the set $\{0,1,\dots, p-2\}$. Now, $i+j+k < 3(p-1)$ and so the only possible cases are $i+j+k \in \{0,p-1,2(p-1)\}$. In the case when the sum vanishes, we are done. If the sum is $2(p-1)$, we can consider the triple obtained by removing $i,j,k$ from $p-1$ (ensured by FLT). In any case, we may assume that $i+j+k=p-1$. Now the congruence equation \begin{align*} (x-a)^i(x-b)^j(x-c)^k=1 \end{align*}is given to have $p-3$ solutions in the domain $F_p[X]$. Assume that $i$ is the largest of the three exponents. This, by the FLT, boils down to the equation \begin{align*} (x-b)^j(x-c)^k-(x-a)^{j+k} =0 \end{align*}having $p-3$ solutions. This equation is not an identity (distinct variables etc) and has degree at most $j+k-1 \le \frac{2(p-1)}{3}-1$ (leading coefficient vanishes). However, by Lagrange's theorem, this shows that $p-3 \le \frac{2(p-1)}{3}-1$, which clearly fails for $p \ge 7$. The result follows.

When you forgot Lagrange's Theorem.

bad solution: Removing the fluff, what we are given is that \[(x-a)^i(x-b)^j(x-c)^k\equiv 1\pmod{p}\]for all $x\in\mathbb{F}_p\setminus\{a,b,c\}$. Also, we might as well take the $i,j,k$ mod $p-1$ because of Fermat's little theorem. Furthermore, we may send $(i,j,k)\to(-i,-j,-k)$ if we wish to becauase $1^{-1}\equiv 1\pmod{p}$. By taking mod $p-1$, we may assume WLOG that $0\le i\le j\le k\le p-2$. We see that $i+j+k=0,p-1,2(p-1)$ since $p-1\mid i+j+k$. If $i+j+k=0$, we're actually done, so for sake of contradiction, suppose not. Now, if $i+j+k=2(p-1)$, then send $(i,j,k)\to(p-1-k,p-1-j,p-1-i)\pmod{p-1}$. So potentially re-ordering again, we may assume $0\le i\le j\le k\le p-2$, and $i+j+k=p-1$. Finally, send $k\to k-(p-1)$, so now we have $0\le|i|,|j|,|k|\le p-2$, $i,j\ge 0$, and $-k=i+j$. Thus, we have \[(x-a)^i(x-b)^j\equiv(x-c)^{i+j}\pmod{p}\]for all $x\not\equiv a,b,c\pmod{p}$. Let $P(x)=(x-a)^i(x-b)^j-(x-c)^{i+j}$.By letting $x=a$, we see that this polynomial is not identically $0$ (here we are working over $\mathbb{F}_p$). Thus, $1\le\deg P\le k-1$, but $P(x)$ has $p-3$ roots (which are $\mathbb{F}_p\setminus\{a,b,c\}$). Therefore, $P(x)$ factors into $p-3$ linear factors, so in fact $p-3\le k-1$. But $k-1\le p-3$, so in fact $p-3=k-1$, so $k=p-2$. It is not hard to see that this means \[(x-a)^i(x-b)^{1-i}(x-c)^{-1}\equiv 1\pmod{p}\]for all $x\not\equiv a,b,c\pmod{p}$. Thus, $(x-a)^i-(x-b)^{i-1}(x-c)\equiv 0\pmod{p}$ for all $x\not\equiv a,b,c\pmod{p}$, so by similar logic to above, we have $i-1\ge p-3$, so $i=p-2$ as well. So we have $i=k=-1$, so $j=2$. Thus, \[(x-b)^2\equiv (x-a)(x-c)\pmod{p}\]for all $x\not\equiv a,b,c\pmod{p}$, so $-2bx+b^2\equiv -(a+c)x+ac\pmod{p}$ for those same $x$ values. If $p>5$, we must then have that $2b\equiv a+c\pmod{p}$ and $b^2\equiv ac\pmod{p}$. Therefore, \[(a-c)^2=(a+c)^2-4ac\equiv 4b^2-4b^2\equiv 0\pmod{p},\]so $a\equiv c\pmod{p}$, which is the desired contradiction. $\blacksquare$

Here's a cleaner finish than using Lagrange. First we may assume that $i, j, k < p-1$ by FLT, so $i+j+k\in \{0, p-1, 2(p-1)\}$. If $i+j+k = 0$, we're done. If $i+j+k = 2(p-1)$, send $i\to (p-1)-i$ (and similarly for $j$ and $k$) so that $i+j+k=p-1$. This does not affect the hypothesis by FLT. Hence we can assume $i+j+k=p-1$. Let $$f(x) = (x-a)^i(x-b)^j(x-c)^k - 1,$$so by hypothesis $f$ vanishes everywhere except at $a, b, c$. This implies $\sum_{x=0}^{p-1} f(x) \equiv -3\pmod p$. However, $f$ is monic with degree $p-1$, and thus $\sum_{x=0}^{p-1} f(x) \equiv -1\pmod p$, contradiction. Here are the details for the last step. If we let $g = f-x^{p-1}$ have degree less than $p-1$, we can write $$\sum_{x=0}^{p-1} f(x) = \sum_{x=0}^{p-1} x^{p-1} + \sum_{x=0}^{p-1} g(x) \equiv -1\pmod p,$$where we are using the fact that if a polynomial $P$ has degree less than $p-1$, then $\sum_x P(x) \equiv 0\pmod p$. Of course this follows from the fact that $$1^k+2^k+\dots +(p-1)^k\equiv 0\pmod p$$if $p-1\nmid k$.

Nice problem! The idea involved is truly magnificent. Here's my solution: If $i \geq p-1$, then replace $i \rightarrow i-(p-1)$, and note that the given condition remains invariant due to Fermat's Little Theorem. So WLOG assume that $0 \leq i,j,k \leq p-2$ (from now on we will take $i,j,k$ modulo $p-1$). Then we have $i+j+k=0,p-1,2(p-1)$. If $i+j+k=0$, then we must have $i,j,k=0$, which is what we needed. If $i+j+k=2(p-1)$, then replace $i \rightarrow p-1-i$, and similarly for others, to get the same condition (again with the help of FLT). So we can assume that $i+j+k=p-1$ and $0 \leq i,j,k \leq p-2$. Now, the given condition is equivalent to saying that $$(x-a)^i(x-b)^j(x-c)^k \equiv 1 \pmod{p} \text{ } \forall x \in \mathbb{F}_p \setminus \{a,b,c\}$$$$\Rightarrow (x-a)^i(x-b)^j \equiv (x-c)^{i+j} \pmod{p} \text{ } \forall x \in \mathbb{F}_p \setminus \{a,b,c\}$$where we use FLT along with the fact that $k=(p-1)-(i+j)$. This means that the polynomial $G(x)=(x-a)^i(x-b)^j-(x-c)^{i+j}$ has $p-3$ roots (where we have $G \in \mathbb F_p[x]$). As degree of $G$ is at most $i+j-1$, so we have $$i+j-1 \geq p-3 \Rightarrow p-2-k \geq p-3 \Rightarrow k \leq 1 \Rightarrow k=0,1$$Now we have the following two cases to deal with:- If $k=0$, then we have $$(x-a)^i(x-b)^j \equiv 1 \pmod{p} \Rightarrow (x-a)^i(x-b)^{p-1-i} \equiv 1 \pmod{p}$$Multiplying both sides of the congruence by $(x-b)^i$, and with the help of FLT, we get that $(x-a)^i \equiv (x-b)^i \pmod{p}$. Again consider the polynomial $H(x)=(x-a)^i-(x-b)^i$, which has $p-3$ roots in $\mathbb F_p[x]$. As $\text{deg}(H) \leq i-1$, so $i-1 \geq p-3$, which gives $i=p-2$ (since $0 \leq i \leq p-2$). But then, working in $\mathbb F_p,$ we have $$(x-a)^{p-2}=(x-b)^{p-2} \Rightarrow (x-a)^{p-1}(x-b)=(x-b)^{p-1}(x-a) \Rightarrow x-b=x-a \Rightarrow a=b \Rightarrow p \mid a-b \rightarrow \text{CONTRADICTION}$$$\text{ }$ Let $k=1$. Then we have $j=p-2-i$, which gives $$(x-a)^i(x-b)^{p-2-i}(x-c) \equiv 1 \pmod{p} \Rightarrow (x-a)^i(x-c) \equiv (x-b)^{i+1} \pmod{p}$$Let $T(x)=(x-a)^i(x-c)-(x-b)^{i+1}$ be a polynomial in $\mathbb F_p[x]$. Then, it has $p-3$ roots, and $\text{deg}(T) \leq i$. This gives $i \geq p-3$, which means that $i=p-3,p-2$ (since $0 \leq i \leq p-2$). If $i=p-2$, then $j=0$, in which case we can arrive at a contradiction in a similar manner as we did for the case $k=0$. So we must have $i=p-3,j=k=1$. This gives $$(x-a)^{p-3}(x-b)(x-c) \equiv 1 \pmod{p} \Rightarrow (x-b)(x-c) \equiv (x-a)^2 \pmod{p} \text{ } \forall x \in \mathbb{F}_p \setminus \{a,b,c\}$$where we multiply both sides by $(x-a)^2$ and use FLT. As this has $p-3$ roots, and $p-3>1$ (since $p>5$), so the identity $(x-a)^2=(x-b)(x-c)$ should be true in $\mathbb F_p$. Thus, we have (working in $\mathbb F_p$) $b+c=2a \text{ and } bc=a^2$. As this is the equality case of the AM-GM inequality, so we must have $b=c$, which contradicts the fact that $p \nmid b-c$. Thus, we arrive at a contradiction in all cases, as desired. $\blacksquare$ EDIT: I forgot to mention that each of the polynomials $G,H,T$ are not identities (which we get on plugging in $a,b,c$ in them). We need this since otherwise we can not apply bounding on their degrees.

pinetree1 wrote: Here's a cleaner finish than using Lagrange. First we may assume that $i, j, k < p-1$ by FLT, so $i+j+k\in \{0, p-1, 2(p-1)\}$. If $i+j+k = 0$, we're done. If $i+j+k = 2(p-1)$, send $i\to (p-1)-i$ (and similarly for $j$ and $k$) so that $i+j+k=p-1$. This does not affect the hypothesis by FLT. Hence we can assume $i+j+k=p-1$. Let $$f(x) = (x-a)^i(x-b)^j(x-c)^k - 1,$$so by hypothesis $f$ vanishes everywhere except at $a, b, c$. This implies $\sum_{x=0}^{p-1} f(x) \equiv -3\pmod p$. However, $f$ is monic with degree $p-1$, and thus $\sum_{x=0}^{p-1} f(x) \equiv -1\pmod p$, contradiction. Here are the details for the last step. If we let $g = f-x^{p-1}$ have degree less than $p-1$, we can write $$\sum_{x=0}^{p-1} f(x) = \sum_{x=0}^{p-1} x^{p-1} + \sum_{x=0}^{p-1} g(x) \equiv -1\pmod p,$$where we are using the fact that if a polynomial $P$ has degree less than $p-1$, then $\sum_x P(x) \equiv 0\pmod p$. Of course this follows from the fact that $$1^k+2^k+\dots +(p-1)^k\equiv 0\pmod p$$if $p-1\nmid k$. You miss the point that $i, j, k$ might be 0, then your claim failed.

Reduce $i,j,k$ modulo $p-1$. If $i+j+k=0$ we are done, and if $i+j+k=2(p-1)$ replace each with $p-1-i,p-1-j,p-1-k$ so we can WLOG assume $i+j+k=p-1$. WLOG $i+j\le 2(p-1)/3$. Clearly $i+j>0$ because otherwise $k$ wasn't reduced. The goal is to show we cannot have \[(x-c)^{i+j}-(x-a)^i(x-b)^j\equiv 0\pmod{p}\]for all $x\not\equiv a,b,c\pmod{p}$. Suppose otherwise. The polynomial has degree at most $i+j-1$ but $p-3$ roots. Note it cannot be zero because then taking $x=c$ yields a contradiction. Otherwise, we get \[p-3\le i+j-1\le 2(p-1)/3-1\iff p-2\le 2(p-1)/3\iff 3p-6\le 2p-2\iff p\le 4,\]which is absurd, thus we are done.

Interesting. Notice by FLT, we can immediately restrict the values of $i,j,k$ to $\{0,1,...,p-2\}$. Hence we have that $i+j+k \le 3(p-2)$, so $i+j+k \in \{0,p-1,2(p-1)\}$.If $i+j+k = 0$, then it is easy to see that we are done. Now we will show that we can reduce the case where $i+j+k = 2(p-1)$ to $i+j+k= p-1$. We can consider two cases now, one where $\max{(i,j,k)} \ge p-1$ and one where $p-1 \ge \max{(i,j,k)} \ge \frac{2(p-1)}{3}$. In the former case, we are able to replace $\max{(i,j,k)}$ with $\max{(i,j,k)} - (p-1)$, and in the latter case replace $\max{(i,j,k)}$ with $(p-1) - \max{(i,j,k)}$. Now we can turn all our attention to the case where $i+j+k = p-1$. Observe that we are essentially operating on $\mathbb{F}_p \setminus \{a,b,c\}$ and have that ${(x-a)}^i{(x-b)}^j{(x-c)}^k = 1$. We desire to show that this isn't possible unless $x=a,b,c$, and we will suppose the contrary. Assume that $\max{(i,j,k)}=i$. Applying FLT once more gives us ${(x-b)}^j{(x-c)}^k - {(x-a)}^{j+k} = 0$, and notice that the leading degree cancels out so the degree of this polynomial is given to be $j+k-1$. Since we have that $\max{(i,j,k)}=i$, we get that $j+k-1 \le \frac{2(p-1)}{3}$. By Lagrange, we have that $p-3 \le \frac{2(p-1)}{3}$, which only holds for $p \le 4$, ridiculous. $\blacksquare$

Reduce $i,j,k$ modulo $p-1$. Since the equation is symmetric we can assume that $i>j>k$(if they were equal then $p-1|i+j+k=3i=3j=3k$ or $p-1|i,p-1|j,p-1|k$,which is what we wanted) If $i+j+k=0$,we will be done so consider the case $i+j+k=2(p-1)$ where we can replace each of $i,j,k$ with $p-i-1,p-j-1,p-k-1$ so we just need to check the case when $i+j+k=p-1$ By FLT,${(x - a)(x - b)(x - c)[(x-b)}^j{(x-c)}^k - {(x-a)}^{j+k}] \equiv 0 \pmod p$,and since this works for all $x$,choose $x$ such that $(x - a)(x - b)(x - c) \equiv r>0 \pmod p$ By Lagrange's theorem and since $p$ is a prime,$f(X) \equiv 0 \pmod p$,has atmost $deg(f)>0$ solutions so $(X-b)^j(X-c)^k \equiv (X-a)^{j+k} \pmod p$ must have atleast $p$ solutions $\pmod p$ which covers all residues so it must work for all integers. Now plugging in $X=b$ and $X=c$ gives $(b-a)^{j+k}$ and $(c-a)^{j+k}$ to be divisible by $p$ Since $p \nmid(b-a),(c-a)$,$j+k=(p-1)r$,a contradiction since $0<j+k<p-1$,or all of $i,j,k \equiv 0 \pmod p$,as required.

Assume that $i, j, k < p-1$. We will show that $i=j=k=0$. Assume otherwise. Then $i+j+k = (p-1)$ or $2(p-1)$; in the former case, replace each $i$ with $p-1-i$, so it suffices to show the former case. WLOG $i>j>k$. We have the reduction $$(x-a)^{j+k}(x-b)(x-c) \equiv (x-b)^{j+1}(x-c)^{k+1}$$for all integers $x$ by Fermat's little theorem. But now, the relevant degree of the polynomial is $j+k+2 < p$, hence $f \equiv 0 \pmod p$ by Lagrange. Then $$(X-b)^j(X-c)^k \equiv (X-a)^{j+k} \pmod p.$$At $X=b$, this implies $a=b$, impossible. Thus $i=j=k=0$.

WLOG $0 \leq i, j, k \leq p-2$ since we can remove $p-1$ from each of $i$, $j$, $k$ repeatedly since it does not affect the quantity in question. If $0 = i = j = k$ we are done so FTSOC assume this is not true. Then $i + j + k < 3(p-1)$ so either $i + j + k = p-1$ or $2p - 1$. If $i + j + k = 2p - 1$ then replace each of $i, j, k$ with $(p-1) - i, (p-1) - j, (p-1) - k$ to get that $i + j + k = p-1$ which is the only case we will be dealing with. Note that $1^k + 2^k + \dots + (p-1)^k \equiv 0\pmod{p}$ unless $p - 1 \mid k$. Then we can count $\sum_{i=1}^{p-1} (x-a)^i(x-b)^j(x-c)^k \pmod{p}$ from our previous claim as $-1 \pmod{p}$. However since $(x-a)^i(x-b)^j(x-c)^k - 1$ vanishes modulo $p$ for all $x \neq a, b, c$ we have that $\sum_{i=1}^{p-1} (x-a)^i(x-b)^j(x-c)^k \equiv p-3 \pmod{p}$. This is a contradiction as $p-1 \equiv p-3 \pmod{p}$ implies $p = 2$, contradiction. So then $i \equiv j \equiv k \equiv 0\pmod{p-1}$.

Storage: Removing the cover; we have: for all $x \in \mathbb F_p /\{a, b, c\}$: $(x-a)^i (x-b)^j (x-c)^k \equiv 1 \equiv (x-a)^{i+j+k} \pmod p$ Assume that: $ i, j, k \le p-1$. If $i+j+k=0$, we are done. If $i+j+k=2(p-1)$, replace $i$ with $p-1-i$, $j$ with $p-1-j$, $k$ with $p-1-k$. Thus assume $i+j+k=(p-1)$. WLOG $i \ge j \ge k$. Thus: $j+k \le \frac{2(p-1)}{3}$. Note that: $(x-b)^j (x-c)^k \equiv (x-a)^{j+k} \pmod p$ has $p-3$ solutions. Thus, by Lagrange's Theorem we must have: $p-3 \le \frac{2(p-1)}{3}$ which leads to contradiction as $p>5$.

Shift $i,j,k$ by $p-1$ to be $<p-1$, then if $i+j+k=2p-2$ consider $p-1-i,p-1-j,p-1-k$ instead. If $i+j+k=p-1$ then $(x-a)^i(x-b)^j(x-c)^k$ is zero on $a,b,c$ and one otherwise, but so is $(x-a)^{p-1}+(x-b)^{p-1}+(x-c)^{p-1}-2$. Now they have the same degree and different leading term, so their difference is a polynomial of degree $p-1$ with $p$ roots $0,1,\dots,p-1$, impossible. Thus $i+j+k=0$ so $i,j,k=0$ which implies $p-1\mid i,j,k$ as desired.