Courtesy of Kalva... n denote the residue of n mod p, so n = 0, 1, 2, ... , or p-1. Thus {na/p} = na/p, and we have that na + nb + nc + nd = 2p for n not a multiple of p. Let ω be a complex pth root of 1. We show first that ω + 2ω2 + 3ω3 + ... + (p-1)ωp-1 = p/(ω - 1). Suppose the sum is S. Then (1 - 2ω + ω2)S = ω - pωp + (p-1)ωp+1 (we need only look at the two lowest and two highest powers - the others all cancel because k - 2(k-1) + (k-2) = 0) = p(ω - 1). Hence S = p/(ω - 1). Take residues a', b', c', d', so that aa' = bb' = cc' = dd' = 1 mod p. Then for any integers m, n we have mnaa' = mn mod p. Hence -ma' na = -mn mod p. Hence ω-ma' na = ω-mn. So na ω-ma' na = na ω-mn. Similarly for b, c, d. Take n not a multiple of p and add the four equations to get: na ω-ma' na + nb ω-mb' nb + nc ω-mc' nc + nd ω-md' nd = 2p ω-mn. As n runs through 1, 2, ... , p-1, each of na, nb, nc, nd runs through a complete set of non-zero residues. If we take m to be not a multiple of p, then so does -mn, so adding the equations for n = 1, 2, ... , p-1, we get na ∑ kω-a'mk + ∑ k ω-b'mk + ∑ kω-c'mk + ∑ kω-d'mk = 2p(ω + ω2 + ... + ωp-1) = -2p. Since ω-a'm is also a complex pth root of 1, we have ∑ kω-a'mk = p/(ω-a'm - 1) and similarly for the other terms. So the equation becomes: 1/(ω-a'm - 1) + 1/(ω-b'm - 1) + 1/(ω-c'm - 1) + 1/(ω-d'm - 1) = -2. Multiplying through by (ω-a'm - 1)(ω-b'm - 1)(ω-c'm - 1)(ω-d'm - 1), expanding and simplifying, we get 2 + ω-a'm-b'm-c'm + ω-a'm-b'm-d'm + ω-a'm-c'm-d'm + ω-b'm-c'm-d'm = ω-a'm + ω-b'm + ω-c'm + ω-d'm + 2ω-a'm-b'm-c'm-d'm (*). Note that this equation is also true for m = 0 (when it is just 5 = 5). Now sum the equations for m = 0, 1, 2, ... , p-1. We have ∑ ωk = p if k is a multiple of p, and 0 otherwise. Obviously the first term ∑ 2 gives 2p and the other terms on the lhs give non-negative sums, so ∑ lhs is at least 2p. The first four terms on the rhs all have zero sum, so the last term must have sum 2p, so a' + b' + c' + d' must be a multiple of p. Thus (*) becomes ωa'm + ωb'm + ωc'm + ωd'm = ω-a'm + ω-b'm + ω-c'm + ω-d'm. Multiply through by ω-a'm and sum for m = 0, 1, 2, ... , p-1. The first term on the lhs has sum p and the others have non-negative sum. The first term on the rhs has zero sum, so one of the others must have positive sum. Hence p divides at least one of (a'+b'), (a'+c'), (a'+d'). Without loss of generality it divides a' + b'. In other words a' + b' = 0 mod p. Multiplying by ab, we get a + b = 0 mod p. Bomb

There are a solution not using primitive root of unity appear in a book of Zuming Feng

spdf wrote: There are a solution not using primitive root of unity appear in a book of Zuming Feng Can you tell the book please? Thanks!

Define R(x) be the remainder obtained upon dividing x by p. Let a' be positive number such that aa' = 1 (mod p), multiply a, b, c, d by a' we have aa', ba', ca', da'. Let $ R(ba') = k_{1}, R(ca') = p-k_{2}, R(da') = p-k_{3}$, WLOG let $ R(ba')\leq R(ca')\leq R(da')$, or $ k_{1}\leq p-k_{2}\leq p-k_{3}$. so $ 1+k_{1}= k_{2}+k_{3}$- ** Now for the sake of contradiction, assume $ k_{3}> 1$. Therefore from ** we see that $ k_{2}, k_{3}< k_{1}$, now WLOG let $ k_{3}\leq k_{2}< k_{1}$ Lemma: $ k_{1}< (p+1)/2$ Proof: By definition $ k_{1}$ is the smallest among $ (k_{1}, p-k_{2}, p-k_{3})$. Assume $ k_{1}\geq (p+1)/2$, then $ R(2k_{1})\leq R(k_{1})-1$, (If$ 2k_{1}-p>k_{1}$, then$ k_{1}>p$ to contradiction) $ R(2(p-k_{2}))\leq R(k_{2})-1$, etc... So let $ r = 2$, $ 2+R(2k_{1})+R(2(p-k_{2}))+R(2(p-k_{3})) < R(1)+R(k_{1})+R(p-k_{2})+R(p-k_{3}) = 2p$, contradiction. Hence the lemma follows. Note that when we multiply 1, k_1, p-k_2, p-k_3 by positive integer r<p, then $ R(rk_{1}) = rk_{1}-[rk_{1}/p]p$ $ R(r(p-k_{2})) = ([rk_{2}/p]+1)p-rk_{2}$ $ R(r(p-k_{3})) = ([rk_{3}/p]+1)p-rk_{3}$ $ R(r)+R(rk_{1})+R(r(p-k_{2}))+R(r(p-k_{3})) = 2p$, from ** we deduce $ [rk_{2}/p]+[rk_{3}/p] = [rk_{1}/p]$- (*) Let $ a$ be the smallest positive integer satisfying $ [ak_{1}/p] = 1$, and simultaneously $ [ak_{2}/p]$ must also be 1 to satisfy (*) (since $ k_{2}\geq k_{3}$). If $ 0 < s < k_{1}$, then either $ a$ or $ a+1$ must be the minimum value for $ m$ so that $ [(s+mk_{1})/p] = 1$ (This is trivial. since $ s < k_{1}$, the minimum for $ m$ is at least $ a$, and, $ s+k_{1}> k_{1}$, so the minimum is at most $ a+1$). Similarly, If $ 0 < s < k_{2}$, then either $ a$ or $ a+1$ must be the minimum value for m so that $ [(s+mk_{2})/p] = 1$. Now let $ m$ be the largest integer value for $ [rk_{1}/p]$, where $ r\leq [p/k_{3}]$. Let $ n$ be the smallest integer sastifying $ [nk_{1}/p] = m$. Since $ n\leq [p/k_{3}]$, $ [nk_{3}/p] = 0$ so by (*), $ [nk_{2}/p] = m$. Claim: If $ t$ is the smallest integer so that $ [tk_{1}/p] = m+1$, then $ t$ is either $ n+a$ or $ n+a+1$. Proof: Note that $ R(nk_{1}) < k_{1}$ (else $ R(nk_{1})-k_{1}\geq 0$ and contradict well-ordering principle ). $ nk_{2}= mp+R(nk_{1})$, so $ [(nk_{2}+(t-n)k_{2})/p] = m+[(R(nk_{1})+(t-n)k_{2})/p]$, minimum for $ t-n$ is either $ a$ or $ a+1$, so miminum for $ t$ is either $ a$ or $ a+1$. Similarly the smallest integer $ t$ satisfying $ [tk_{2}/p] = m+1$ is either $ n+a$ or $ n+a+1$. Now consider setting $ r$ to be $ [p/k_{3}]+1$ (since $ k_{3}> 1$ by assumption, $ [p/k_{3}]+1 < p$.) So $ [k_{3}r/p] = 1$, and the only way to satisfy (*) is if $ [k_{1}r/p] = m+1$, and $ [k_{2}r/p] = m$, which implies $ r = n+a$ (if $ r = n+a+1$, both $ [k_{1}r/p], [k_{2}r/p] = m+1$ and (*) is violated). Consider $ r = [p/k_{3}]+2 = n+a+1$, $ [rk_{2}/p] = m+1$, and By Lemma $ R((n+a)k_{1})+k_{1}< 2k_{1}< p$, so $ [(n+a+1)k_{1}/p] = [m+1+(R((n+a)k_{1})+k_{1})/p ] = m+1$, Similarly, $ [rk_{3}/p] = [1+(R((n+a)k_{3})+k_{3})/p ] = 1$ , and (*) is violated, contradiction. It follows $ k_{3}= 1$, so $ R(da') = p-1$ => $ R(da')+R(aa') = p$ => $ p|(a'd+aa')$ => $ p|(a+d)$ QED

I have no idea if this is useful.

please read this for the solution...

Solution by dysfunctionalequations and me.

My solution to this problem as follows: From the assumption we have: $\frac{r(a+b+c+d)}{p}=2+[\frac{ra}{p}]+[\frac{rb}{p}]+[\frac{rc}{p}]+[\frac{rd}{p}]$ is an integer. And we get $a+b+c+d$ is divisible by $p$. Now, we work with modulo $p$. Let $a+b=x,a+c=y,a+d=z$ leads $c+d=-x,b+d=-y,b+c=-z$ We have: $2xyz=(a+b)(a+c)(a+d)+(b+c)(c+d)(d+b)$ or $2xyz=a^2(a+b+c+d)+a(bc+cd+db)+bcd+(b+c)(c+d)(d+b)$ But $a=-(b+c+d)$ so $a(bc+cd+db)+bcd+(b+c)(c+d)(d+b)=0$. From which we get $p$ divides $2xyz$ but $p>2$ is prime, so: or $p|x$ or $p|y$ or $p|z$ And we assume $p|x$. hence $a+b$ and $c+d$ are divisible by $p$. Q.E.D With the solution above, we can replace 2 by 1 or 3. A very nice problem!

LoveMath4ever wrote: Now, we work with modulo $p$. Let $a+b=x,a+c=y,a+d=z$ leads $c+d=-x,b+d=-y,b+c=-z$ We have: $2xyz=(a+b)(a+c)(a+d)+(b+c)(c+d)(d+b)$ That's not true, because LHS is actually 0?

@bomb Your complex solution is very very beautiful.But I still have something to understand.I can't come up with such beautiful complex solution,Idon't know how it was thinked about.Can you say something about your thought when you were solving the problem?

bomb wrote: Courtesy of Kalva... n denote the residue of n mod p, so n = 0, 1, 2, ... , or p-1. Thus {na/p} = na/p, and we have that na + nb + nc + nd = 2p for n not a multiple of p. Let ω be a complex pth root of 1. We show first that ω + 2ω2 + 3ω3 + ... + (p-1)ωp-1 = p/(ω - 1). Suppose the sum is S. Then (1 - 2ω + ω2)S = ω - pωp + (p-1)ωp+1 (we need only look at the two lowest and two highest powers - the others all cancel because k - 2(k-1) + (k-2) = 0) = p(ω - 1). Hence S = p/(ω - 1). Take residues a', b', c', d', so that aa' = bb' = cc' = dd' = 1 mod p. Then for any integers m, n we have mnaa' = mn mod p. Hence -ma' na = -mn mod p. Hence ω-ma' na = ω-mn. So na ω-ma' na = na ω-mn. Similarly for b, c, d. Take n not a multiple of p and add the four equations to get: na ω-ma' na + nb ω-mb' nb + nc ω-mc' nc + nd ω-md' nd = 2p ω-mn. As n runs through 1, 2, ... , p-1, each of na, nb, nc, nd runs through a complete set of non-zero residues. If we take m to be not a multiple of p, then so does -mn, so adding the equations for n = 1, 2, ... , p-1, we get na ∑ kω-a'mk + ∑ k ω-b'mk + ∑ kω-c'mk + ∑ kω-d'mk = 2p(ω + ω2 + ... + ωp-1) = -2p. Since ω-a'm is also a complex pth root of 1, we have ∑ kω-a'mk = p/(ω-a'm - 1) and similarly for the other terms. So the equation becomes: 1/(ω-a'm - 1) + 1/(ω-b'm - 1) + 1/(ω-c'm - 1) + 1/(ω-d'm - 1) = -2. Multiplying through by (ω-a'm - 1)(ω-b'm - 1)(ω-c'm - 1)(ω-d'm - 1), expanding and simplifying, we get 2 + ω-a'm-b'm-c'm + ω-a'm-b'm-d'm + ω-a'm-c'm-d'm + ω-b'm-c'm-d'm = ω-a'm + ω-b'm + ω-c'm + ω-d'm + 2ω-a'm-b'm-c'm-d'm (*). Note that this equation is also true for m = 0 (when it is just 5 = 5). Now sum the equations for m = 0, 1, 2, ... , p-1. We have ∑ ωk = p if k is a multiple of p, and 0 otherwise. Obviously the first term ∑ 2 gives 2p and the other terms on the lhs give non-negative sums, so ∑ lhs is at least 2p. The first four terms on the rhs all have zero sum, so the last term must have sum 2p, so a' + b' + c' + d' must be a multiple of p. Thus (*) becomes ωa'm + ωb'm + ωc'm + ωd'm = ω-a'm + ω-b'm + ω-c'm + ω-d'm. Multiply through by ω-a'm and sum for m = 0, 1, 2, ... , p-1. The first term on the lhs has sum p and the others have non-negative sum. The first term on the rhs has zero sum, so one of the others must have positive sum. Hence p divides at least one of (a'+b'), (a'+c'), (a'+d'). Without loss of generality it divides a' + b'. In other words a' + b' = 0 mod p. Multiplying by ab, we get a + b = 0 mod p. Bomb https://mks.mff.cuni.cz/kalva/usa/usoln/usol993.html

Zhero wrote: The polynomial $S(r) = P(ra) + P(rb) + P(rc) + P(rd)$ is congruent to 2 mod $p$ for $p-1$ values. Since the degree of $S$ is $p-2$, $S-2$ must be the zero polynomial mod $p$. Since $S$ has degree $p-2$, the coefficient of $r^{p-2}$in $P(r)$ is nonzero mod $p$. Since $P(ra) + P(rb) + P(rc) + P(rd) - 2$ is the zero polynomial, we must have that $a^{p-2} + b^{p-2} + c^{p-2} + d^{p-2} \equiv 0 \pmod{p}$, or $\frac{1}{a} + \frac{1}{b} \equiv -\frac{1}{c} - \frac{1}{d} \pmod{p}$. In particular, it isn't clear why this is true: Zhero wrote: Since $P(ra) + P(rb) + P(rc) + P(rd) - 2$ is the zero polynomial, we must have that $a^{p-2} + b^{p-2} + c^{p-2} + d^{p-2} \equiv 0 \pmod{p}$

Delray wrote: Zhero wrote: The polynomial $S(r) = P(ra) + P(rb) + P(rc) + P(rd)$ is congruent to 2 mod $p$ for $p-1$ values. Since the degree of $S$ is $p-2$, $S-2$ must be the zero polynomial mod $p$. Since $S$ has degree $p-2$, the coefficient of $r^{p-2}$in $P(r)$ is nonzero mod $p$. Since $P(ra) + P(rb) + P(rc) + P(rd) - 2$ is the zero polynomial, we must have that $a^{p-2} + b^{p-2} + c^{p-2} + d^{p-2} \equiv 0 \pmod{p}$, or $\frac{1}{a} + \frac{1}{b} \equiv -\frac{1}{c} - \frac{1}{d} \pmod{p}$. In particular, it isn't clear why this is true: Zhero wrote: Since $P(ra) + P(rb) + P(rc) + P(rd) - 2$ is the zero polynomial, we must have that $a^{p-2} + b^{p-2} + c^{p-2} + d^{p-2} \equiv 0 \pmod{p}$ collab with janabel Actually the fact that S(r) is the zero polynomial means that S(r) = 0, which means everything about it equals 0. Also since it is a p-2 degree polynomial, we can just take the p-2 degree terms and have that ra^{p-2} + rb^{p-2} + rc^{p-2} + rd^{p-2} = 0. Setting r =1 results in the desired equation.

vsathiam wrote: Actually the fact that S(r) is the zero polynomial means that S(r) = 0, which means everything about it equals 0. Wow much rigorous.

Solution from Twitch Solves ISL: First of all, we apparently have $r(a+b+c+d) \equiv 0 \pmod p$ for every prime $p$, so it automatically follows that $a+b+c+d \equiv 0 \pmod p$. By scaling appropriately, and also replacing each number with its remainder modulo $p$, we are going to assume that \[ 1 = a \le b \le c \le d < p. \]We are going to prove that $d = p-1$, which will solve the problem. Claim: For each integer $r = 1, 2, \dots, p-1$ we have \[ 2(r-1) = \left\lfloor \frac{rb}{p} \right\rfloor + \left\lfloor \frac{rc}{p} \right\rfloor + \left\lfloor \frac{rd}{p} \right\rfloor. \]Proof. By plugging in $r=1$ to the given we have $a+b+c+d=2p$. Now, we have \[ 2 = \sum_{\text{cyc}} \left( \frac{ra}{p} - \left\lfloor \frac{ra}{p} \right\rfloor \right) \]and since $a+b+c+d=2p$ the conclusion follows. $\blacksquare$ We vaguely outline the approach now, before giving a formalization. Imagine the interval $[0,1]$. One by one, for each $r = 1, 2, 3, \dots, p-1$, we mark the fractions with denominator $r$ on this number line; the resulting pictures may be better known as Farey fractions. At each step, we can place the three numbers $b/p$, $c/p$, $d/p$ into one of the resulting sub-intervals. Our goal is to show that $d/p$ is always in the rightmost interval, while $b/p$ and $c/p$ are always to the right of symmetrically mirrored points. An example of a possible diagram is shown below (not to scale). [asy][asy] unitsize(11cm); usepackage("amsmath"); real h = 0.05; draw( (0,0)--(1,0) ); void tick(string s, real x, real t, pen p) { label(s, (x,-t), dir(-90), p); draw((x,-t)--(x,t), p); } tick("$0$", 0, h, black); tick("$1$", 1, h, black); tick("$\frac{1}{2}$", 1/2, h, black); tick("$\frac{1}{3}$", 1/3, 0.7*h, blue); tick("$\frac{2}{3}$", 2/3, 0.7*h, blue); tick("$\frac{1}{4}$", 1/4, 0.5*h, deepgreen); tick("$\frac{3}{4}$", 3/4, 0.5*h, deepgreen); tick("$\frac{1}{5}$", 1/6, 0.3*h, grey); tick("$\frac{2}{5}$", 2/5, 0.3*h, grey); tick("$\frac{3}{5}$", 3/5, 0.3*h, grey); tick("$\frac{4}{5}$", 5/6, 0.3*h, grey); dot("$\boxed{\frac{b}{p}}$", (0.3,0), dir(90), red); dot("$\boxed{\frac{c}{p}}$", (0.72,0), dir(90), red); dot("$\boxed{\frac{d}{p}}$", (0.95,0), dir(90), red); [/asy][/asy] In symbols, It will be enough to prove the following. Claim: For each $r = 1, 2, \dots, p-2$ we have $\frac{r-1}{r} < \frac dp < 1$. Equivalently, for each $r = 1, 2, \dots, p-2$ we have $\left\lfloor \frac{rb}{p} \right\rfloor + \left\lfloor \frac{rc}{p} \right\rfloor = r-1$. Proof. Assume this is not true and take the minimal counterexample $r > 1$. Then evidently \[ r-1 > \left\lfloor \frac{rd}{p} \right\rfloor \ge \left\lfloor \frac{(r-1)d}{p} \right\rfloor = r-2. \]Now, we have that \[ 2(r-1) = \left\lfloor \frac{rb}{p} \right\rfloor + \left\lfloor \frac{rc}{p} \right\rfloor + \underbrace{\left\lfloor \frac{rd}{p} \right\rfloor}_{= r-2}. \]Thus $\left\lfloor \frac{rb}{p} \right\rfloor > \left\lfloor \frac{(r-1)b}{p} \right\rfloor$, and $\left\lfloor \frac{rc}{p} \right\rfloor > \left\lfloor \frac{(r-1)b}{p} \right\rfloor$. An example of this situation is illustrated below with $r = 7$ (not to scale). [asy][asy] unitsize(11cm); usepackage("amsmath"); real h = 0.05; draw( (0,0)--(1,0) ); void tick(string s, real x, real t, pen p) { label(s, (x,-t), dir(-90), p); draw((x,-t)--(x,t), p); } tick("$0$", 0, h, black); tick("$1$", 1, h, black); tick("$\frac{1}{2}$", 1/2, h, black); tick("$\frac{1}{3}$", 1/3, 0.7*h, blue); tick("$\frac{2}{3}$", 2/3, 0.7*h, blue); tick("$\frac{1}{4}$", 1/4, 0.5*h, deepgreen); tick("$\frac{3}{4}$", 3/4, 0.5*h, deepgreen); tick("$\frac{1}{5}$", 1/6, 0.3*h, grey); tick("$\frac{2}{5}$", 2/5, 0.3*h, grey); tick("$\frac{3}{5}$", 3/5, 0.3*h, grey); tick("$\frac{4}{5}$", 5/6, 0.3*h, grey); tick("$\frac{1}{6}$", 1/8, 0.3*h, grey); tick("$\frac{5}{6}$", 7/8, 0.3*h, grey); dot("$\boxed{\frac{b}{p}}$", (0.31,0), dir(90), red); dot("$\boxed{\frac{c}{p}}$", (0.725,0), dir(90), red); dot("$\boxed{\frac{d}{p}}$", (0.92,0), dir(90), red); tick("$\dfrac{2}{7}$", 2/7, 1.5*h, red+1); tick("$\dfrac{5}{7}$", 0.7, 1.5*h, red+1); tick("$\dfrac{6}{7}$", 0.9, 1.5*h, red+1); tick("$\dfrac{7}{8}$", 0.95, 2*h, orange+1); tick("$\dfrac{3}{8}$", 0.375, 2*h, orange+1); tick("$\dfrac{6}{8}$", 0.75, 2*h, orange+1); [/asy][/asy] Right now, $\frac bp$ and $\frac cp$ are just to the right of $\frac ur$ and $\frac vr$ for some $u$ and $v$ with $u+v=r$. The issue is that the there is some fraction just to the right of $\frac bp$ and $\frac cp$ from an earlier value of $r$, and by hypothesis its denominator is going to be strictly greater than $1$. It is at this point we are going to use the properties of Farey sequences. When we consider the fractions with denominator $r+1$, they are going to lie outside of the interval they we have constrained $\frac bp$ and $\frac cp$ to lie in. Indeed, if right now $\frac ur < \frac bp < \frac st$, then Farey theory says the next fraction inside the interval $[\frac uv, \frac st]$ is $\frac{u+s}{v+t}$, and since $t > 1$, we have $v+t > r+1$. In other words, we get an inequality of the form \[ \frac ur < \frac bp < \text{something} \le \frac{u+1}{r+1}. \]The same holds for $\frac cp$ as \[ \frac vr < \frac cp < \text{something} \le \frac{v+1}{r+1}. \]Finally, \[ \frac dp < \frac{r-1}{r} < \frac{r}{r+1}. \]So now we have that \[ \left\lfloor \frac{(r+1)b}{p} \right\rfloor + \left\lfloor \frac{(r+1)c}{p} \right\rfloor + \left\lfloor \frac{(r+1)d}{p} \right\rfloor \le u + v + (r-1) = 2r-1 \]which is a contradiction. $\blacksquare$ Now, since \[ \frac{p-3}{p-2} < \frac dp \implies d > \frac{p(p-3)}{p-2} = p - 1 - \frac{2}{p-2} \]which for $p > 2$ gives $d = p-1$.

This was fun... I didn't really use anything besides floors