It's trivial with Gauss and Jacobi sums, reduces to $p-4\sqrt{p} \geq 1$ and triangle inequality. http://people.reed.edu/~jerry/361/lectures/lec11.pdf seems to develop the needed stuff, and even has a couple of similar looking examples.

Problems of this type have been discussed in detail in the paper T. Mautsch and G. J. Woeginger, The Sum of a Cube and a Fourth Power, Crux Mathematicorum 34 (2008), pp 358-361. This paper shows (among other results of this type) that such values $a$ and $b$ exist for every prime $p$ and every integer $r$ ($0\le r<p$) with the single exception of $p=11$ and $r=7$.

An alternative way to do this problem without high powered results (ok, Jacobi sums aren't that complicated but are pretty hard to motivate if one has never seen them before) is to count the number of times where $a^2 + b^5 \equiv c^2 + d^5 \pmod{p}$. Obviously WLOG $p \equiv 1 \pmod{10}$. Note that we have $(a-c)(a+c) \equiv d^5 - b^5 \pmod{p}$, so for fixed $b,d$ which are not equal there are exactly $p-1$ pairs of $a,c$ which which make the two equal. If $b=d$ then there are exactly $2p-1$ pairs. Thus over the $p^4$ quadruples of $(a,b,c,d)$, exactly \[(p^2-5p + 4) \cdot (p-1) + (5p-4)\cdot (2p-1) = p^3 + 4p^2 - 4p\] quadruples satisfy $a^2 + b^5 \equiv c^2 + d^5 \pmod{p}$. Denote $a_n$ as the number of solutions to $a^2 + b^5 \equiv n \pmod{p}$. Clearly we must have \[\sum_{n=0}^{p-1} a_n^2 = p^3 + p^2 - p\] and \[\sum_{n=0}^{p-1} a_n = p^2\] Let $p = 10k+1$. If the problem is false, there exists at least $k$ values of $n$ such that $a_n = 0$ (because if $u$ is not obtainable, then $u \cdot x^{10}$ is not either for all $x \neq 0$). Thus there exists a set $S \subset \mathbb{Z_p}$ such that $|S| = 9k+1$ and $\sum_{x \in S} a_x = p^2$. Then by Cauchy: \[\sum_{x \in S} a_x^2 \ge \frac{p^4}{9k+1}\] Now, clearly $\sum_{x \in S} a_x^2 = p^3 + 4p^2 - 4p$. Its not hard (but slightly ugly) to check that $\frac{p^4}{9k+1} > p^3 + 4p^2 - 4p$ for $k \ge 4$, a contradiction as $p > 100$, so the result follows.

dinoboy wrote: An alternative way to do this problem without high powered results (ok, Jacobi sums aren't that complicated but are pretty hard to motivate if one has never seen them before) is to count the number of times where $a^2 + b^5 \equiv c^2 + d^5 \pmod{p}$. Obviously WLOG $p \equiv 1 \pmod{10}$. Note that we have $(a-c)(a+c) \equiv d^5 - b^5 \pmod{p}$, so for fixed $b,d$ which are not equal there are exactly $p-1$ pairs of $a,c$ which which make the two equal. If $b=d$ then there are exactly $2p-1$ pairs. Thus over the $p^4$ quadruples of $(a,b,c,d)$, exactly \[(p^2-5p + 4) \cdot (p-1) + (5p-4)\cdot (2p-1) = p^3 + 4p^2 - 4p\] quadruples satisfy $a^2 + b^5 \equiv c^2 + d^5 \pmod{p}$. Denote $a_n$ as the number of solutions to $a^2 + b^5 \equiv n \pmod{p}$. Clearly we must have \[\sum_{n=0}^{p-1} a_n^2 = p^3 + p^2 - p\] and \[\sum_{n=0}^{p-1} a_n = p^2\] Let $p = 10k+1$. If the problem is false, there exists at least $k$ values of $n$ such that $a_n = 0$ (because if $u$ is not obtainable, then $u \cdot x^{10}$ is not either for all $x \neq 0$). Thus there exists a set $S \subset \mathbb{Z_p}$ such that $|S| = 9k+1$ and $\sum_{x \in S} a_x = p^2$. Then by Cauchy: \[\sum_{x \in S} a_x^2 \ge \frac{p^4}{9k+1}\] Now, clearly $\sum_{x \in S} a_x^2 = p^3 + 4p^2 - 4p$. Its not hard (but slightly ugly) to check that $\frac{p^4}{9k+1} > p^3 + 4p^2 - 4p$ for $k \ge 4$, a contradiction as $p > 100$, so the result follows.

I'm not sure if this solution is entirely correct since 36 is way smaller than 100.

This is also trivial with the Hasse-Weil bound, which says that the number of points $N_p$ on a curve mod p with genus $g$ satisfies \[|N_p - p|\le 2g\sqrt{p}\] (I am omitting the point at infinity) By the genus-degree formula, $g < \binom{4}{2} = 6$. Therefore, we are done for all primes $p > 144$. Manually checking all the primes less than $36$ is not very hard since one can eliminate everything that isn't $1\pmod5$. Therefore, we just have to check $101, 131$, which doesn't take much time. EDIT: This can actually be improved a lot. Since $y^2 = x^5 + k$ is a hyperelliptic curve, it has genus $\left\lfloor \frac{5- 1}{2}\right\rfloor = 2$, so we have that by the method above, for all $p > 16$, there exist solutions $\pmod{p}$.

The second solution to this problem in the official solutions also show that every prime except 11 satisfies the problem.

http://imo-official.org/problems/IMO2012SL.pdf if you want the official solutions.

This is the standard approach but whatever. The problem is trivial unless $5\mid p-1$, so let $p=5k+1$. Suppose $a^2+b^5\equiv c^2+d^5\pmod{p}$. Then $t\equiv a^2-c^2\equiv b^5-d^5\pmod{p}$. If $t=0$ then there are $1+4(p-1)/2=2p-1$ choices of $a,c$. Similarly, there are $1+25k=5p-4$ choices of $b,d$. If $t\ne 0$ then consider the values $a=\frac{s+t/s}{2}$ for $s\ne 0$ modulo $p$. It is clear that \[s+t/s\equiv u+t/u\implies (s-u)\equiv t\cdot (s-u)/us\implies t\equiv us\]modulo $p$ so there are $2(p-1)/2=p-1$ choices of $a,c$. Let the number of choices of $b,d$ with $b^5-d^5\equiv t$ be $f(t)$. Now we count $(2p-1)(5p-4)+\sum_{t\ne 0}(p-1)f(t)$ different such overlaps. The latter sum is clearly \[(2p-1)(5p-4)+(p-1)(p^2-5p+4)=10p^2-13p+4+p^3-6p^2+9p-4=p^3+4p^2-4p.\]For each $r\pmod p$, let $g(r)$ be the number of choices of $a,b$ with $a^2+b^5\equiv r\pmod p$. Then \[\sum_r g(r)^2 = p^3+4p^2-4p.\]Each fixed $a^2+b^5$ (where $a^2$ and $b^5$ are fixed) occurs $5$ times if $a=0$ and $b\ne 0$, $2$ times if $a\ne 0$ and $b=0$, $1$ time if $a=b=0$, and $10$ times if $a,b\ne 0$. Moreover, the case of something occurring $5$ times only happens in $k$ cases for $a$, the case of something occurring $2$ times only happens in $(p-1)/2$ cases for $b$, and the last case only happens with $r=0$, for which the solutions are those to $a^2\equiv (-b)^5$, which has $2\cdot5\cdot k/2+1=p$ solutions. So $\sum_{r\ne 0} g(r)^2=p^3+3p^2-4p$. At this point, remark that if a particular $r$ is not in the range of $a^2+b^5$, then so must be all $rg^{10}$ for $g\in\mathbb F_p$: that is, the number of nonzero elements of the range of $a^2+b^5$ is a multiple of $k/2$. Remark that by definition of $g(r)$, $\sum_{r\ne 0} g(r)=p^2-p$. Thus if some $r$ is not in the range of $a^2+b^5$ (and clearly $r\ne 0$) then \[(p^3+3p^2-4p)\cdot\frac{9(p-1)}{10}\ge (p^2-p)^2\]by Cauchy-Schwarz. Equivalently, \[(p^2+3p-4)\cdot \frac{9}{10}\ge (p^2-p)\iff p+4\ge \frac{10}{9}(p-1)=\frac{10}{9}p-\frac{10}{9}\iff \frac p9\le \frac{46}{9}.\]But $p\ge 100$, so we win.

We uploaded our solution https://calimath.org/pdf/ISL2012-N8.pdf on youtube https://youtu.be/B69Nvz88nzQ.

If $p \not \equiv 1 \pmod{5}$ the result is trivial. Otherwise let $\psi$ be the legendre symbol and $\chi$ a quintic character. Note that the number of solutions to $a^2 + b^5 \equiv r$ is \[ \sum_{a+b \equiv r} (1 + \psi(a))(1 + \chi(b) + \chi(b^2) + \chi(b^3) + \chi(b^4)). \]Define $J_r(\chi, \psi) = \sum_{t \in \mathbb F_p} \chi(t) \psi(r - t)$, and $r$ nonzero. Then by a changing of variables $t \to rt$ one notes that $J_r(\chi, \psi) = \chi\psi(r) J_1(\chi, \psi)$. Moreover $\sum_{a \in \mathbb F_p} \chi^k(a) = 0$ for $k = 1, 2, 3$ and $4$. We also have $\sum_{a \in \mathbb F_p} \psi(a) = 0$. Thus when expanding the sum we get \[ p + J_r(\chi, \psi) + J_r(\chi^2, \psi) + J_r(\chi^3, \psi) + J_r(\chi^4, \psi). \]Note that $\chi^k\psi$ is nontrivial. As such we have \[ J_r(\chi^k, \psi) = \chi\psi(r) \cdot \frac{g(\chi^k)g(\psi)}{g(\chi^k \psi)}, \]where $g(\chi) = \sum_{t = 0}^p \chi(t)e^{\frac{2\pi i t}{p}}$ is the standard Gauss sum. From known computations one obtains that $|J_r(\chi^k, \psi)| = \sqrt{p}$, whence \[ p + J_r(\chi, \psi) + J_r(\chi^2, \psi) + J_r(\chi^3, \psi) + J_r(\chi^4, \psi) \ge p-4\sqrt{p}. \]Since $p > 100$ one notes that this quantity is greater than $0$, done.

Essentially same as #12. But we will do this problem instead. Quote: Let $p$ be a prime and $r$ be an integer such that $0 \leq r <p$. Find all such pairs $(r,p)$ for which there exists no integers $(a,b)$ such that $p \mid a^2+b^5-r$. Call a prime $p$ good iff for all integers $r$, there exists such $a$ and $b$. Obviously $p=2$ is good so assume $p$ is odd prime. Claim: All primes $p \not \equiv 1 \pmod 5$ is good. Proof: This is obvious as the map $b \mapsto b^5$ in $\mathbb{F}_p^{\times}$ just permutes the elements. $\square$. Hence from now on assume $b \equiv 1 \pmod 5$. Claim: All such primes $p \geq 17$ are good. Proof: See that $r=0$ case is obvious and hence assume $r > 0$. Fix $L$ be a Dirichlet character of order $2$ (so basically the Legendre symbol) and $\chi$ be a Dirichlet character of order $5$ (make sure that their absolute value is $1$). What follows for now, $\chi_1$ and $\chi_2$ will denote any random Dirichlet characters (of any order). Denote the Gauss Sum of a character as $\tau(\chi_1)$. And denote the Jacobi Sum of two characters as $J(\chi_1,\chi_2)$. Now it is well known that \[N(a^2+b^5=r)=\sum_{\chi_1^2=\chi_2^5=\epsilon} \chi_1(r)\chi_2(r) J(\chi_1,\chi_2)\]And because of orthogonality relations and the fact that $J(\epsilon,\epsilon)=p$, we get that the value is equal to \begin{align*} & \text{ }p+L\chi(r)J(L,\chi)+\dots+L\chi^4(r) J(L,\chi^4) \\ = & \text{ } p+L \chi(r) \frac{\tau(L)\tau(\chi)}{\tau(L\chi)} + \dots + L\chi^4(r) \frac{\tau(L) \tau(\chi^4)}{\tau(L \chi^4)} \\ \geq & \text{ } p-4 \sqrt{p} \end{align*}See that we got the second sum because of the well known fact that \[J(\chi_1,\chi_2)=\frac{\tau(\chi_1) \tau(\chi_2)}{\tau(\chi_1 \chi_2)}\]and the last sum is due to good old triangle inequality and the fact that $|\tau(\chi_1)|=|\tau(\chi_2)|=\sqrt{p}$. And hence the result follows. $\square$. So we just have to check for primes $p < 17$ and $p \equiv 1 \pmod 5$ which is just $p=11$ and now just case-bash and see that the only solution is $\boxed{(r,p)=(7,11)}$.

Claim 1: We just want to show that $a^2$ $\equiv $ $r-b^5$ (mod $p$), which is stating that $r-b^5$ is a quadratic residue for some $b$. We know that $x^n$ gives $\frac{n}{(n,p-1)}$ residues mod $p$. Hence $p=10k+1$ or $b^5$ gives all the residues mod $p$ so we are done. Claim 2: Using the Legandre's symbols, we now know that $(_{p}^{r-b^5})=-1$. Therefore $\forall b$: $(r-b^5)^\frac{p-1}{2}\equiv-1$ (mod $p$) Claim 3: $1\equiv(r-1^5)^\frac{p-1}{2}+...(r-(p-1)^5)^\frac{p-1}{2}\equiv\sum(r-b^5)^\frac{p-1}{2}$ $b^{5m}$$\equiv$$\sum^{5k}_{2k|i+m}(-1)^{5i}C^i_{5k} r^{5k-i}$ because $1^n+2^n+...+(p-1)^n \equiv 0 $ (mod $p$) when $(p-1,n)=1$ and $-1$ else. Then the only thing you are left to do is to take $m=1$ and $m=2$ which will result in something like this: $r^{2k}\equiv-\frac{(2k-1)!(3k+1)!}{(k+1)!(4k-1)!}$ (mod $p$) $r^{2k}\equiv-\frac{(2k-2)!(3k+2)!}{(k+2)!(4k-2)!}$ (mod $p$) Which, after subtracting one from the otherwill give something like $2k(5k+1)\equiv0$ (mod $p$) which is not true because $p=10k+1$ so we are done. .

cursed wrote: $1\equiv(r-1^5)^\frac{p-1}{2}+...(r-(p-1)^5)^\frac{p-1}{2}\equiv\sum(r-b^5)^\frac{p-1}{2}$ $b^{5m}$$\equiv$$\sum^{5k}_{2k|i+m}(-1)^{5i}C^i_{5k} r^{5k-i}$ Can you please explain this line? Like what is $k$ and $m$ here?