Perhaps only p=2 works. I just thought it could be written in the way that p(p-1)=x³+1=(x+1)(x²-x+1) and do some discussions......

There is actually another prime that works (the only other one I found under $10,000$). $$37^2 - 37 - 1 = 11^3 = 1331$$

We have $-1 = a^3 \mod p$ so $a \mod p =p-1 $ or $6 | p-1$.

Here is the idea. Let $p^2-p-1=x^3$. Then $p(p-1)=(x+1)(x^2-x+1)$ so $p\mid x+1$ or $p\mid x^2-x+1$. If $p\mid x+1$ then we expect $x\sim p^{\frac{2}{3}}$ so $x$ will be too large. If $p\mid x^2-x+1$, then let $x^2-x+1=kp$ so $p-1 = k(x+1)$. As we expect $k\sim x^{\frac{1}{2}}$, it's a good idea to bound $x$ as a quadratic polynomial in $k$. With some computation, we can reach the conclusion.

This one looks like a problem from BMO

this was 2012 Belarus TST 2.1 proposed by S. Mazanik & I. Voronovich

To solve this topic, you can rewrite the equation like : $$ p^2 -p - 1= x^3 \implies p(p-1) = X(X^3 - 3X +3) $$ With $ X= x+1 $ , $x$ an integer. The resolution become such a game consisting on an exchange of factor between $X$ and $X^2 -3X + 3$ to obtain $p(p-1)$.The equation become the search of integers with $b < a$ in the conditions $a| X^2- 3X +3; b| X$ so that: $$\frac{b}{a} .(X^2 - 3X +3)-\frac{a}{b}X = \pm1$$. It is not difficult to determine that: $6| p-1$ and $6|X$ when $p$ is $odd$. To solve the problem we have to distinguish cases: --> $b|a \implies b=1$ --> $2\leq b < a$ and $b\nmid a$ Expecially here, it is possible to determine maximum and minimum value admissible for $X$. For more informations see here: edemafa wrote: $ P R E L I M I N A R Y $ ---> We start with these transformations : $y^3 = x(x+6) \implies$ $y^3 -9 = (x+3)^2$ $\implies (y+2)(y^2 -2y+4) = (x+2)(x+4)$ $(1)$ Here we can observe that $-2\leq y $ ---> $\Delta_x(t)$ is a discriminant with the unknow $x$ and parameter $t$ $ S E C O N D$ $ M E T H O D $ $(1) \implies (y+2)(y^2 -2y+4) = (x+2)(x+4)$ Noting $S = y+2$ and $X = x+ 2 $ we obtain $ (y+2)(y^2 -2y+4) = S (S^2 - 6S +12)$ and $ (1) \implies S (S^2 - 6S +12) = X (X+2)$ Note that $X = -1 \pm \sqrt{9 + (S-2)^3}$ $(2)$ The problem become a game consisting on the arrangement of factors of $S$ and $S^2 - 6S +12$ to get X so that we get the relationship: $ (1) \implies S (S^2 - 6S +12) = X (X+2)$ Firstly, it is obvious that $S^2 - 6S +12 > 0$ and $ S^2 - 6S +12 \geq S$ with egality when $S = 3 $ $or$ $4$ which are not solutions. So we can assume for $S$ solutions: $ S^2 - 6S +12 > S$ Let $a$ and $b$ with $a \mid S^2 - 6S +12 $ and $b \mid S$. It is clear that $b \leq a$. We distinguish two cases: <--> $b \mid a \implies b=1$ <--> $b \nmid a \implies 2 \leq b < a$ The aim is to solve the equation: $ b\frac {S^2 - 6S + 12 }{a} - a\frac {S}{b} = \pm2$ $ C A S E$ $1$ $( b = 1) $ $ \frac {S^2 - 6S + 12 }{a} - aS = \pm2 \implies S^2 - (a^2 +6)S +12 - \pm 2a =0$ $\Delta_S(a) = (a^2 +6)^2 \pm 8a -48 $ $(3)$ $\boxed{\frac {S^2 - 6S + 12 }{a} - aS = 2}$ $(3) \implies \Delta_S(a) = (a^2 +6)^2 + 8(a - 6) $ Let $h$ an integer define by: $\Delta_S(a) = (a^2 + 6)^2 + 8(a - 6)$ $= (a^2 + 6 +2h)^2 $ $\implies h( a^2 + 6 + h ) = 2 (a - 6)$ Observe that the sign of $h$ depend on $(a-6)$ . It is easy to chech that : $a \geq 7 \implies a^2 +6 > 2 (a-6)$ and it is impossible to get $h$ an integer. Thus we have to check $a \in [1;6]$. The discriminant is a square for: $(a; \Delta_S) = (1; 3^2), (6; 42^2)$ ** $ a = 6 ;$ $\Delta_S = 42^2 \implies h=0$ $$S = \frac{6^2 +6 \pm 42}{2} = 0 ; 42$$ By $(2)$ we get: - $ S = 0 \implies X = -1 \pm \sqrt{9 + (0 - 2)^3}$ $= -2; 0$ - $ S = 42 \implies X = -1 \pm \sqrt{9 + (42 - 2)^3}$ $= -254 ; 252$ ** $ a = 1 ;$ $\Delta_S = 3^2 \implies h=-5$ $or$ $-2$ $$S = \frac{1^2 +6 \pm 3}{2} = 2 ; 5$$ By $(2)$ we get: - $ S = 2 \implies X = -1 \pm \sqrt{9 + (2 - 2)^3}$ $= -4; 2$ - $ S = 5 \implies X = -1 \pm \sqrt{9 + (5 - 2)^3}$ $= -7 ; 5$ $\boxed{\frac {S^2 - 6S + 12 }{a} - aS = -2}$ $(3) \implies \Delta_S(a) = (a^2 +6)^2 - 8(a + 6) $ Let h an integer define by: $\Delta_S(a) = (a^2 + 6)^2 - 8(a + 6) = (a^2 + 6 -2h)^2 $ $\implies h( a^2 + 6 - h ) = 2 (a + 6)$ $(4)$ Observe that $a \geq 7 \implies h > 0$. If $a = 10$ : $(4) \implies h(106 - h) = 32$ Very obvious that it is absurd to determinate $h$ so that $h \leq 32$ and $ h \geq 74$. And if $a \geq10$ , $a^2 +6 $ grow faster thant $2(a+6)$ so $h$ can't be an integer. By checking all natural numbers less tha $10$ we find: $(a; \Delta_S) = (2; 6^2)$ ** $ a = 2 ; \Delta_S = 6^2 \implies h= 2$ or $8$ $$S = \frac{2^2 +6 \pm 6}{2} = 2 ; 8$$ By $(2)$ we get: - $ S = 2 $ yet done - $ S = 8 \implies X = -1 \pm \sqrt{9 + (8 - 2)^3}$ $= -16; 14$ All the results put together gives: $(S;X) = (0; 0), (0; -2), (2; -4), (2; 2), (5; -7), (5; 5), (8; -16), (8; 14), (42; -254), (42; 252) \implies$ $\boxed{(y,x) = (-2;-4), (-2;-2), (0;-6), (0;0), (3;-9),}$ $\boxed{(3,3), (6;-18), (6;12), (40; -256), (40; 250)}$ Now let demonstrate that for the second part, we have not new solutions. This demonstration is very long. $ C A S E$ $2: $ $ (2 \leq b < a ;$ $ b\nmid a) $ Here we have: $ b\frac {S^2 - 6S + 12 }{a} - a\frac {S}{b} = \pm2$ $\implies S^2 - ( \frac {a^2}{b^2} + 6)S + 12 - \pm 2\frac {a}{b} = 0$ $\Delta_S = (\frac {a^2}{b^2} + 6)^2 - 48 \pm 8\frac {a}{b}$ $\boxed{b\frac {S^2 - 6S + 12 }{a} - \frac{a}{b}.S = 2}$ $\Delta_S = (\frac {a^2}{b^2} + 6)^2 + 8(\frac{a}{b} - 6)$ Let $h$ a $rationnal$ $number$ so that : $\Delta_S = ( \frac {a^2}{b^2} + 6)^2 + 8(\frac{a}{b} - 6)$ $ = ( \frac {a^2}{b^2} + 6 + 2h)^2 $ It is clear that the sign of $h$ depend on the sign of $ a - 6b$. We have: $(\frac {a^2}{b^2} + 6)^2 + 8(\frac{a}{b} - 6)= ( \frac {a^2}{b^2} + 6 + 2h)^2 $ $\implies h(a^2 + 6b^2 + hb^2) = 2.b.(a-6b)$ $(5)$ And : $S = \frac {1}{2} . [(\frac {a^2}{b^2} + 6) \pm (\frac {a^2}{b^2} + 6 + 2h)]$ $ S = -h $ $or$ $\frac {a^2}{b^2} + 6 + h$ $(6)$ So $h$ is an $integer$ or a $pure$ $rational$ <=====> $h$ is an integer $(5) \implies h(a^2 + 6b^2 + hb^2) = 2.b.(a-6b)$ So $b \nmid a \implies b \mid h$ $(7)$ Easy to observe that : * $a \geq 6b \implies a^2 + (h+6) b^2 \geq 2b(a-6b)$ So $h$ can not be an integer. Contradiction with our hypothesis ( $h$ integer) *$a = 6b \implies S = 0 ; 42$ * $a \leq 6b \implies h \leq 0$ $So$ $a^2 + (h+6) b^2 \geq 0$ and with : $ (6b)^2 > a^2 > -(h + 6) b^2\implies h > -42$ $h > -42 ;$ $ S = -h ;$ $S =0 ; 42 $ $\implies \boxed{ S \in \{0, 1 . . . 40, 41, 42} /$ This result gives an interval to search $S$ by computer. Let continue with our mathematical method. Considering $k$ an positive integer so that: $ h(a^2 + 6b^2 + hb^2) = 2b(a-6b) \implies \frac {h}{b} = \frac {2a - 12b}{a^2 + ub^2} = -k$ with $u=h+6$ $(8)$ $ \frac {2a - 12b}{a^2 + ub^2} = -k $ $ \implies ka^2 + 2a -12b + kb^2u = 0$ $\Delta'_a = 1 - k ( -12b + kb^2u)$ $= -uL^2 + 12 L + 1 $ with $L = kb > 0$ $(9)$ ** Let demonstrate that if $u > 0 \implies h \in [-6; -1] $ we have no solution. - $h = -1 :$ $ b\mid h \implies h \leq -2$ so no solution. - $h = -2 :$ Taking $b = 2$ and plugging in $(5)$ gives $a= 2$ which is not possible because $ b \mid a$ - $h = -3 :$ Taking $b = 3$ and plugging in $(5)$ doesn't give solution for $a$. - $h = -4 :$ Taking $b = 2$ $or$ $4$ and plugging in $(5)$ doesn't give solution for $a$. - $h = -5 :$ Taking $b = 5$ and plugging in $(5)$ gives $a= 5$ which is not possible because $ b \mid a$ - $h = -6 :$ Taking $b = 2$ $or$ $3$ $or$ $6$ and plugging in $(5)$ doesn't give solution for $a$. So it is done. ** Let determinate good solutions when $u < 0 \implies h \in [-36; -7] $ . $\Delta'_a = -uL^2 + 12 L + 1 $ $\implies \Delta'_a = (k_1L + 1)^2$ $or$ $(k_1L - 1)^2$ , $k_1$ a positive integer If $\Delta'_a = (k_1L + 1)^2$ : $a = \frac{-1 + (k_1L + 1)}{k} \implies \frac {a}{b} = \frac{k_1L }{L} $ (because $L = k.b$). Contradiction with $b \nmid a$. If $\Delta'_a = (k_1L - 1)^2 .$ When $L=2 ,$ $ 2k_1 - 1 = 2 (k_1 - 1) +1$ $\implies \frac {a}{b} = k_1 - 1$ which is a contradiction. Then $L \geq 3$ and we can write: $(k_1L - 1)^2 = -uL^2 + 12 L + 1 \implies L = \frac{2k_1 + 12}{k_1^2 + u}$ with $k_1 \neq \sqrt {-u}$ With this kind of fonction, the condition on $L$ and $u \implies k_1 \leq 7$. Taking in account also that $k_1 \geq \sqrt {-u}$, here is the value to be verify by $k_1$ in function of $u,$ an negative integer : - if $ u \in [-2, -1] \implies k_1 = 2; 3$ - if $ u \in [-5, -4] \implies k_1 = 3$ - if $ u \in [-8, -6] \implies k_1 = 3; 4$ - if $ u \in [-13, -9] \implies k_1 = 4$ - if $ u \in [-15, -14] \implies k_1 = 4; 5$ - if $ u \in [-16, -23] \implies k_1 = 5$ - if $ u = -24 \implies k_1 = 5 ; 6$ - if $ u \in [-35, -25] \implies k_1 = 6$ - if $ u = -36 \implies k_1 = 7$ To get an positive integer $L \geq 3$, here are the list of $u$: $ u \in \{-24, -24, -15, -14, -12, -11, -8, -7, -6, -3, -2 \}$ With $S= -h$ and $u = h+6 \implies S = -u +6$ $\implies S \in \{ 8, 9, 12 , 13, 14, 17, 18, 20, 21, 29, 30 \}$. Only $S=8$ gives solutions and this case is yet studied. <=====> $h$ is a pure rational number $ S= \frac {a^2}{b^2} + 6 + h$ $(6) \implies \exists h' \in \mathbb{Z} / h = \frac{h'}{b^2}$ and $\frac {a^2 + h'}{b^2} \in \mathbb{N}*$ $(5) \implies h(a^2 + 6b^2 + hb^2) = 2.b.(a-6b)$ $\implies h' \frac {a^2 + 6 b^2 + h'}{b^2} = 2b(a-6b) $ $\implies \frac {a^2 + 6 b^2 + h'}{b^3} = \frac {2(a-6b)}{h'} = k$ As $b\nmid a \implies b\nmid h' \implies k$ is an integer. Also $h'$ has the same sign like $h$, it has the same sign like $a - 6b$ gives $k$ is a positive integer. So we can solve, with our new contraint : $ \frac {a^2 + 6 b^2 + h'}{b^3} = k $ $\implies ka^2 + 2a + 6kb -k^2b^3 -12b =0 $ by using $ 2(a - 6b) = h'.k$ $\Delta'_a = 1 - k(6kb -k^2b^3 -12b)$ $\Delta'_a = (kb)^3 - 6(kb)^2 + 12 (kb) +1$ ** If with $k_1 \in \mathbb{N}*$: $\Delta'_a = (k_1L + 1)^2$ , posing $L = kb$, so $\frac {a}{b} = \frac {-1 + k_1L + 1}{L} = k_1$ contraduction with $b \nmid a$ ** If with $k_1 \in \mathbb{N}*$: Posing $L = kb$, $\Delta'_a = (k_1L - 1)^2$ $= L^3 - 6L^2 + 12L +1$ $\implies L^2 - (k_1 + 6)L + 12 + 2k_1 = 0$ $ \Delta_L = (k_1 + 6)^2 - 4 ( 12 + 2k_1)$ $ \Delta_L = (k_1 + 2)^2 - 16$ So $k_1 = 2$ ou $k_1 = 3$. -- $k_1 = 2 \implies L= 4 $ >> With $L = 4 ; b = 2 \implies a = 3$ $ \implies h' = -9 \implies S = 6$ >> With $L = 4 ; b = 4 \implies a = 6$ $\implies h' = -36 \implies S = 6$ -- $k_1 = 3 \implies L= 3; 6$ >> With $L = 3 ; b = 3 \implies a = 7$ $\implies h' = -22 \implies S = 9$ >> With $L = 6 ; b = 3 \implies a = 8$ $\implies h' = -10 \implies S = 12$ >> With $L = 6 ; b = 6 \implies a = 16$ $\implies h' = -40 \implies S$ don't exist >> With $L = 6 ; b = 2 \implies a = 16$ $\implies h'$ is not a negative integer, contradiction so no solution. If $S = 6$ $or$ $12$ we can't find $X$ by $(2)$. $\boxed{b\frac {S^2 - 6S + 12 }{a} - \frac{a}{b}.S = 2}$ USING THE SAMES STEPS WILL GIVES US SOLUTIONS ALL YET FOUND. AND THE WORK IS DONE.

cooljoseph wrote: There is actually another prime that works (the only other one I found under $10,000$). $$37^2 - 37 - 1 = 11^3 = 1331$$ You are true. By demonstration, it suffice to check $ [0 ; 12] $ for $ X=x+1 $ . Generally, if $ x^3 - l^3 $ is the second of the equation, we can write : $x^3 + l^3 = X(X^2 - U.X +P)$ with $X= x+ l$. The value solution of $X \in [0, P^2 +P] , $ $P$ a natural number. I think it is in general case.

Let $p^2-p-1=x^3$ then $p(p-1)=(x+1)(x^2-x+1)$ Case 1: $p|x+1$ then $x^2-x+1|p-1\Rightarrow x+1\ge p$ and $p-1\ge x^2-x+1$ so $x\ge p-1\ge x^2-x+1\Rightarrow 0\ge (x-1)^2\Rightarrow x=1\Rightarrow p=2$ This gives us the solution $(p,x)=(2,1)$ Case 2: $p|x^2-x+1$. Suppose $x^2-x+1=pk$ then $p-1=(x+1)k$ so $x^2-x+1=pk=[(x+1)k+1]k=k^2x+k^2+k\Rightarrow x^2-x(k^2+1)-(k^2+k-1)=0$ $\Delta = k^4+6k^2+4k-3$ but $(k^2+5)^2>k^4+6k^2+4k-3>(k^2+1)^2$ and by $(\mod{2})$ we get $k^4+6k^2+4k-3=(k^2+3)^2\Rightarrow k=3\Rightarrow x=-1,11$. $x=-1$ gives us $p=1,0$ Contradiction. $x=11$ gives the solution $(p,x)=(37,11)$

BarisKoyuncu wrote: Let $p^2-p-1=x^3$ then $p(p-1)=(x+1)(x^2-x+1)$ Case 1: $p|x+1$ then $x^2-x+1|p-1\Rightarrow x+1\ge p$ and $p-1\ge x^2-x+1$ so $x\ge p-1\ge x^2-x+1\Rightarrow 0\ge (x-1)^2\Rightarrow x=1\Rightarrow p=2$ This gives us the solution $(p,x)=(2,1)$ Case 2: $p|x^2-x+1$. Suppose $x^2-x+1=pk$ then $p-1=(x+1)k$ so $x^2-x+1=pk=[(x+1)k+1]k=k^2x+k^2+k\Rightarrow x^2-x(k^2+1)-(k^2+k-1)=0$ $\Delta = k^4+6k^2+4k-3$ but $(k^2+5)^2>k^4+6k^2+4k-3>(k^2+1)^2$ and by $(\mod{2})$ we get $k^4+6k^2+4k-3=(k^2+3)^2\Rightarrow k=3\Rightarrow x=-1,11$. $x=-1$ gives us $p=1,0$ Contradiction. $x=11$ gives the solution $(p,x)=(37,11)$ Very proud of your method. But for the case 2, the discriminant allow so much solutions that every thing that follow after is not rigorous. We can write by modulo (2) also : $k^4+6k^2+4k-3=(k^2-49)^2\Rightarrow k=7.$ I think you must avoid the discriminant way like this because you forget forcely a lot of cases.