$\mathbb Z^+ \times \left\{ 1 \right\}$ and $\left(\left( 11\mathbb Z \pm 1\right) \cap \mathbb Z^+ \right)\times \left\{ 11 \right\}$, sir. We consider only the case $b>1$. Suppose that $P$ is a prime power dividing $b$. Evidently $P$ is not even. Then there exists an $n$ for which \[ n^5 \equiv (n+1)^5 \equiv (n-1)^5 \pmod{P}. \]Therefore, \[ 0 \equiv (n+1)^5 + (n-1)^5 - 2n^5 = 20n^3+10n \pmod{P}. \]Since $n \not\equiv 0 \pmod{p}$ for obvious reasons, we obtain that \[ n^2 \equiv -\tfrac{1}{2} \pmod{p}. \]Then, \begin{align*} 0 &\equiv 2((n+1)^5 - (n-1)^5) \pmod{P} \\ &= 20n^4+40n^2+4 \\ &\equiv 20 \cdot -\frac{1}{4} + 40 \cdot -\frac{1}{2} + 4 \pmod{P} \\ &= -11. \end{align*}This implies $P = 11$. Since $P$ can be any prime power dividing $b$, this forces $b=11$. In other words, $b \in \left\{ 1,11 \right\}$. Finally, observe that \[ 3^5 \equiv 4^5 \equiv 5^5 \equiv 1 \pmod{11} \]and \[ (-3)^5 \equiv (-4)^5 \equiv (-5)^5 \equiv -1 \pmod{11} \]are the only such triples modulo $11$. Therefore the solution set is the one claimed above.

Darn v_Enhance can solve and type faster than I can solve and type apparently. Clearly $b=1$ works so take $b > 1$. Note that we simply need $n^5, (n+1)^5, (n+2)^5$ to be congruent modulo $b$. It suffices to find for which prime powers this is possible by CRT. We will show in fact $p=11$ is the only prime which works. Now, if a prime $p$ works clearly $p \neq 5$ so $p \equiv 1 \pmod{5}$. Now note that \[(n+2)^5 - 2(n+1)^5 + n^5 = 10(n+1)(2n^2 + 4n + 3)\] Thus it follows we require either $n+1 \equiv 0 \pmod{p}$ or $2n^2 + 4n + 3 \equiv 0 \pmod{p}$. It is clear that $n \equiv -1 \pmod{p^k}$ does not work, so $2n^2 + 4n + 3 \equiv 0 \pmod{p^k}$. Performing the division algorithm of $2n^2 + 4n + 3$ into $(n+1)^5 - n^5$ (here we use $p \neq 2$ as $p \equiv 1 \pmod{5}$), we get $\frac{-11}{4}$ must be $0 \pmod{p^k}$. Thus $b=11$ is the only possible solution as the only prime power dividing it can be $11$. As $6^5 \equiv 7^5 \equiv 8^5 \pmod{11}$, it follows $p=11$ works. Thus a full solution set is $a = \text{Anything}, b = 1$ and $a = \text{Any positive integer 1 or 10 modulo 11}, b = 11$ (to see why $a = \pm 1$ work just consider $n=3$ and $n = 6$).

$b|(n-1)^5+a$ $b|n^5+a$ $b|(n+1)^5+a$ for some $n$. From here imeditaly folows that $b$ is odd. Then: $b| (n+1)^5-n^5 = 5n^4+10n^3+10n^2+5n+1$ and $b| n^5-(n-1)^5 = 5n^4-10n^3+10n^2-5n+1$ so $b|20n^3 +10n$ (substract previously formulas) and $b|10 n^4+20n^2+2$ and since $b$ is odd we have $b|10n^3 +5n$ and $b|5 n^4+10n^2+1$ From here we have $b|(10 n^4+20n^2+2)-n(10n^3 +5n)= 15n^2+2$ Now $b| 3(10n^3 +5n) -2n(15n^2+2) =11n$ Finally: $b| 11( 15n^2+2) -15(11n) = 22$. So $b\in \{1,11\}$ Obivusly for $b= 1$ every $a$ works. For $b=11$ every $a$ such that $11|a^2-1$ works, since by Fermat $n^{10}\equiv 1 (\mod 11)$

dinoboy wrote: Darn v_Enhance can solve and type faster than I can solve and type apparently. Probably more so the typing, although it does help to use $n-1,n,n+1$ rather than $n,n+1,n+2$. dinoboy wrote: It says positive integers $a,b$ You've got to be kidding me. Fixed, along with a large number of edits for LaTeX reformatting (something about this forum which annoys me a lot.) Number1 wrote: For $b=11$ every $a$ such that $11|a^2-1$ works, since by Fermat $n^{10}\equiv 1 (\mod 11)$ I think you do need to check that the run exists explicitly? Otherwise it might just be runs of length $1$ and $2$. Nothing important, though. By the way, just use \pmod{11} rather than (\mod{11}).

Suppose that the consecutive integers are $n-1,n,n+1$. Then $(n-1)^5 \cong n^5 \cong (n+1)^5(mod$ $b)$. This means that $b|(n+1)^5 +(n-1)^5 -2n^5 = 20n^3 +10n$. If $n\cong 0(mod$ $b)$, consider any prime $p|b$. Then $p|a$ and $p|a+1$, contradiction. So $b \nmid n$. Then using the notion of inverse modulos with the theory of fractions, $n^2 \cong \frac{-1}{2}(mod$ $p^m)$, where $p^m|b, p\in \mathbb{P}$. Then $0\cong (n+1)^5 -(n-1)^5 = 10n^4 +20n^2 +2 \cong \frac{10}{4} + 20\frac{-1}{2} +2 \cong \frac{-11}{2}(mod$ $p^m)$. Then $p^m|11$. So $b=11$. Now it is enough to show that there exist such integers for $b=11$. Observe that $n^5 \cong \left(\frac{n}{11}\right)$. For this $3^5 \cong 4^5 \cong 5^5 \cong 1(mod$ $11)$ and similarly $6^5 \cong 7^5 \cong 8^5 \cong -1(mod$ $11)$. So we can let $n\cong 4(mod$ $11)$ and $a\cong 10(mod$ $11)$ or $n\cong 8(mod$ $11)$ and $a \cong 1(mod$ $11)$. Other than that the case $b=1$ is obvious.

v_Enhance wrote: Find all positive integers $a$ and $b$ for which there are three consecutive integers at which the polynomial \[ P(n) = \frac{n^5+a}{b} \]takes integer values. We claim that the solutions are $(a,b)=(k,1), (l, 11),$ where $k$ is any integer and $l$ is any integer such that $11|l \pm 1.$ These work, and now we will show that these are the only solutions. Firstly assume $b>1,$ and say that $P(n-1), P(n), P(n+1) \in \mathbb{Z}.$ Then \begin{align} P(n+1)+P(n-1)-2P(n) \in \mathbb{Z} \implies b|20n^3+10n \\ P(n+1)-P(n-1) \in \mathbb{Z} \implies b|10n^4+20n^2+2 \end{align} Hence, $b|2(10n^4+20n^2+2)-n(20n^3+10n)=30n^2+4$ and $b|2n(30n^2+4)-3(20n^3+10n)=-22n.$ Thus, $b|22n,$ which we will refer to as $(3).$ Claim: If $p|b,$ then $p=11.$ Further, $v_{11}(b) \le 1$ Proof: If $p=2,$ then $2|n^5+a$ and $2|(n+1)^5+a$ which implies $2|(n+1)^5-n^5,$ which is not possible. Next assume $p>2.$ Then we must have $p \nmid 2n,$ otherwise $(2) \implies p|2,$ absurd. So $p \nmid 2n, p|22n \implies p=11$ Now $11 \nmid 2n \implies 11 \nmid n$ and so $v_{11}(b) \le v_{11}(22n) =1,$ and the claim has been proven. $\square$ Thus, $b=11$ as $b>1$ by our assumption. Now we have proven that $11 \nmid n$ and so $(1) \implies 11|2n^2+1.$ Thus $n^2 \equiv 5 \Leftrightarrow n \in \{4, 7\} \pmod{11}.$ Now since $(4-1)^5 \equiv 4^5 \equiv (4+1)^5 \equiv 1 \pmod{11}$ as well as $(7-1)^5 \equiv 7^5 \equiv (7+1)^5 \equiv -1 \pmod{11},$ hence the solutions are indeed the claimed ones. $\blacksquare$

Can't we just bash this with Euclidean algorithm? The answer is $b=1$ and any $a$ ; $b=11$ and $a \equiv \pm 1 \pmod{11}$. Suppose $b \mid n^5 + a, (n+1)^5 + a, (n+2)^5 + a$. By orders stuff we get all primes factors of $b$ must be $1$ modulo $5$. Observe, \begin{align*} & \qquad ((n+1)^5 - n^5,(n+2)^5 - n^5) \\ &= (5n^4 + 10n^3 + 10n^2 + 5n + 1, 10n^4 + 40n^3 + 80n^2 + 80n + 32) \\ &= (5n^4 + 10n^3 + 10n^2 + 5n+1, 20n^3 + 60n^2 + 70n + 30) \\ &= (5n^4 + 10n^3 + 10n^2 + 5n+1,2n^3 + 6n^2 + 7n + 3) \\ &\textcolor{blue}{=} ~ (10n^4 + 20n^3 + 20n^2 + 10n + 2, 2n^3 + 6n^2 + 7n + 3) \\ &= (-10n^3 + -15n^2 - 5n + 2, 2n^3 + 6n^2 + 7n + 3) \\ &= (10n^3 + 15n^2 + 5n - 2, 2n^3 + 6n^2 + 7n + 3) \\ &= (-15n^2 - 30n - 17, 2n^3 + 6n^2 + 7n + 3) \\ &\textcolor{blue}{=}~ (15n^2 + 30n + 17, 30n^3 + 90n^2 + 105n + 45) \\ &= (15n^2 + 30n + 17, 30n^2 + 71n + 45) \\ &= (15n^2 + 30n + 17, 11n + 11) \\ & \textcolor{red}{=} ~ (15n^2+ 30n + 17,11) \\ &= (4n^2 + 8n + 28,11) \\ &= (n^2 + 2n + 7,11) \\ &= ((n+1)^2 + 6, 11) \\ &= \begin{cases} 11 \qquad & \text{if } n \equiv 3,8 \pmod{11} \\ 1 \qquad & \text{otherwise} \end{cases} \end{align*}where the colored equalities do not directly follow by Euclidean algorithm (still, the blue ones follow by all prime factors of $b$ are $1$ mod $5$ and $b$ is odd ; the red one follows by $(15n^2+30n+17,n+1) = (2,n+1)$). So if $b \ne 1$, then we must have $b = 11$. Moreover, $a$ must be $-3^5$ or $-8^5$ modulo $11$, completing the proof. $\blacksquare$

Does this work? When $b=1$ this polynomial always takes integer values so all $(a,b)=(k,1)$ works. When $b=11,$ \[\{0,1,2,\dots,10\}^5\equiv \{0,1,-1,1,1,1,-1,-1,-1,1,-1\}\]so $a=11k+1,11k+10$ for nonnegative integers $k$ works with $b=11.$ Thus, $(11k+1,11)$ and $(11k+10)$ are also valid. We show that these are the only possible values. $~$ If the statement is true for $n-1,n,n+1$ then $P(n+1)-P(n)$ must be an integer, so $b|5n^4+10n^3+10n^2+5n+1$ and $b|5n^4-10n^3+10n^2-5n+1$ $~$ Thus, $b|10n^4+20n^2+2$ and $b|20n^3+10n.$ It is clear that $b$ must be odd, so $b|10n^4+5n^2.$ Hence, $b|15n^2+2.$ Thus, $b|30n^4+4n^2$ and $b|30n^4+15n^2.$ We conclude that $b|11n^2.$ $~$ Note that $\gcd(5n^4+10n^3+10n^2+5n+1,n)=1$ so $\gcd(b,n)=1.$ Thus, $b|11,$ as desired.

We claim the answer is $(1,a)$ and $(11,11\ell\pm 1).$ The problem is equivalent to finding $(a,b)$ such that there exists an $n$ modulo $b$ such that $$(n-1)^5\equiv n^5\equiv (n+1)^5\equiv -a\pmod{b}.$$Notice \begin{align*}20n^3+10n\equiv(n+1)^5+(n-1)^5-2n^5\equiv 0&\pmod{b},\\5n^4+10n^2+1\equiv (n+1)^5-(n-1)^5\equiv 0&\pmod{b}.\end{align*}Hence, $$11n\equiv (4n^4+10n^2+1)(-4n)+(10n^3+5n)(2n^2+3)\equiv 0\pmod{b}.$$Suppose $d=\gcd(n,b)>1.$ Then, $d\mid n\mid n^5+(n+1)^5$ so $d\mid (n+1)^5,$ which is absurd as $d\nmid n+1.$ Hence, $b\mid 11.$ Clearly, if $b=1,$ all $a$ work. If $b=11,$ then $a^2\equiv n^{10}\equiv 1\pmod{b}$ by FLT, so $a\equiv\pm 1\pmod{b}.$ If $a\equiv 1\pmod{b},$ then $n=4$ suffices as a construction as $3^5\equiv 4^5\equiv 5^5\equiv 1\pmod{11}.$ If $a\equiv -1\pmod{b},$ then $n=7$ works as a construction since $6^5\equiv 7^5\equiv 8^5\equiv -1\pmod{11}.$ $\square$

This is just gcd bash. Let the three consecutive numbers be $n-1,n,n+1$ where $n\geq 2$. Since $b\mid a+(n-1)^5,a+n^5,a+(n+1)^5$, we have $b\mid n^5-(n-1)^5,(n+1)^5-n^5$. Let’s consider their gcd. \begin{align*} \gcd(n^5-(n-1)^5,(n+1)^5-n^5) &= \gcd(5n^4+10n^3+10n^2+5n+1,5n^4-10n^3+10n^2-5n+1) \\ &= \gcd(5n^4+10n^3+10n^2+5n+1,20n^3+10n) \\ &= \gcd(5n^4+10n^3+10n^2+5n+1,2n^3+n) \\ &= \gcd(5n^4+10n^2+1,2n^3+n) \\ &= \gcd(5n^4+10n^2+1,2n^2+1) \\ &= \gcd(10n^4+20n^2+2,10n^2+5) \\ &= \gcd(15n^2+2,10n^2+5) \\ &= \gcd(5n^2-3,10n^2+5) \\ &= \gcd(5n^2-3,11) \end{align*}Thus, the gcd is either $1$ or $11$ so $b=1,11$. If $b=1$ then obviously any three consecutive positive integers work so $a$ can be anything. If $b=11$ then the gcd must be $11$. This means that $11\mid 5n^2-3$ so $n\equiv 4,7\pmod{11}$. If $n\equiv 4\pmod{11}$ then $$(n-1)^5\equiv n^5\equiv (n+1)^5\equiv 1\pmod{11}.$$Hence, $11\mid a+1$ so $a\equiv 1\pmod{11}$. If $n\equiv 7\pmod{11}$ then $$(n-1)^5\equiv n^5\equiv (n+1)^5\equiv -1\pmod{11}.$$Hence, $11\mid a-1$ so $a\equiv 1\pmod{11}$. We conclude that all possible pairs are $$\boxed{(a,b)=(k,1),(11k-10,11),(11k-1,11)}$$for all positive integers $k$.

What :no_mouth: . I claim that the answers are $(b,a)=(1,\mathbb N)$ and $(11,\mathbb N\equiv \left\{1,10\right\}\pmod{11})$. Clearly the first one satisfies and for $b=11$, for $a\equiv 1\pmod{11}$ pick $n=7$ and for $a\equiv 10\pmod{11}$ pick $n=4$. By $(x,y)$, I denote the GCD of the two numbers. Now for the proof, note that $(b,2)=1$ and $(b,5)=1$ which also gives $(b,n)=1$. Then literally expand the $(n-1)^5\equiv n^5\equiv (n+1)^5\pmod{b}$ and then some bash gives that $b\in\left\{1,11\right\}$. Now for the uniqueness for $b=11$, just note that $x^5\equiv\left\{0,\pm 1\right\}\pmod{11}$ which just follows from FLT and finishes the problem.

The condition essentially says that there are three consecutive $n$ such that $$n^5+a\equiv 0\pmod b.$$Consider letting them be $n-1,n,n+1.$ Then, $$n^5+5n^4+10n^3+10n^2+5n+1+a\equiv 0\pmod b$$$$n^5+a\equiv0\pmod b$$$$n^5-5n^4+10n^3-10n^2+5n-1+a\equiv 0\pmod b$$By subtracting these, we obtain $$b\mid 5n^4+10n^3+10n^2+5n+1$$and $$b\mid 5n^4-10n^3+10n^2-5n+1,$$so $$b\mid \gcd(5n^4+10n^3+10n^2+5n+1, 5n^4-10n^3+10n^2-5n+1).$$Then. $$\gcd(5n^4+10n^3+10n^2+5n+1, 5n^4-10n^3+10n^2-5n+1)$$$$=\gcd(20n^3+10n,5n^4-10n^3+10n^2-5n+1)$$Note that the second argument is always 1 mod $10n$, so it is always relatively prime to $10n$ and thus we can take that factor out: $$=\gcd(2n^2+n,5n^4-10n^3+10n^2-5n+1).$$Since the first argument is odd, we are allowed to double the second argument: $$=\gcd(2n^2+1,10n^4-20n^3+20n^2-10n+2)$$$$=\gcd(2n^2+1,-20n^3+15n^2-10n+2)$$$$=\gcd(2n^2+1,15n^2+2)$$$$=\gcd(2n^2+1,30n^2+4)$$$$\gcd(2n^2+1,-11).$$This is 11 if $n\equiv4,7\pmod{11}$ and 1 otherwise. Hence, $b$ is either 1 or 11. If $b$ is 1, then any $a$ works, and if $b$ is $11$, then $a$ is either 1 or 10 mod 11.

Headsolved while Dottedcaculator, Bluelinfish, IAmtheHazard were playing $6$ card poker. First consider the case of nontrivial $p^k$. Claim: If $n$ is a prime power then $p = 11, k = 1$. Proof. This means that \[ n^5 - 5n^4 + 10n^3 - 10n^2 + 5n - 1 \equiv n \equiv n^5 + 5n^4 + 10n^3 + 10n^2 + 5n + 1 \pmod{b}. \]from which it follows that $10n^4 + 20n^2 + 2 \equiv 10n^3 + 5n \equiv 0 \pmod{b}$ As such, it must follow that either $n(10n^2 + 5) \equiv 0 \pmod{p^k}$. If $n^2 \equiv - \frac{1}{2}$, it then follows that $\frac{10}{4} - 10 + 2 \equiv 0 \pmod{p^k}$, from which it follows that $p = 11$ and $k = 1$. In the second case, $n \equiv 0$, which can't hold. $\blacksquare$ When $b = 11$, this holds for $n \equiv 4 \pmod{11}$, giving the solution set of $\frac{1}{11}(n^5 + 10 + 11k)$. When $b = 1$, then $P(n) = n^5 + a$ is the solution set.

bad problem lol. This is a sketch, see remark for why Gcd/euclid bash that $(n-1)^2,n^5,(n+1)^5$ all being equal mod $b$. We get $b=1,11$ are the only values of $b$. For $b=1$, any $a$ works. For $b=11$, we can manually determine which $a$ work by bashing out cases. Remark: Lost my writeup for this question b/c taking papers from home to school is too hard. Thats why there are not many details at all in this oops. Luckily I was able to remember sol!

I claim that all solutions are $\boxed{(a, b) = (k, 1), (11k \pm 1, 11)}$, for $k \in \mathbb{Z}$. Let $Q(n) = n^5 + a$. Then observe the following: \begin{align*} b \mid Q(n + 1) + Q(n - 1) - 2Q(n) &\implies b \mid 20n^3 + 10n \\ b \mid Q(n + 1) - Q(n - 1) &\implies b \mid 10n^4 + 20n^2 + 2 \\ b \mid 2(10n^4 + 20n^2 + 2) - n(20n^4 + 10n) &\implies b \mid 30n^2 + 4 \\ b \mid 2n(30n^2 + 4) - 3(20n^2 + 10n) &\implies b \mid -22n. \end{align*}Now \[\gcd(Q(n + 1) - Q(n - 1), n) = 1 \implies \gcd(b, n) = 1 \implies b \mid 22.\]Now we split into four cases: Case 1: ($b = 1$). Then any integer value of $a$ works. Case 2: ($b = 2$). Then we cannot have three consecutive fifth powers with the same remainder modulo $2$, so no solution here exists. Case 3: ($b = 11$). Here observe that \[1 \equiv 3^5 \equiv 4^5 \equiv 5^5 \pmod {11} \implies \boxed{a \equiv -1 \pmod {11}}. \]Also similarly observe \[-1 \equiv (-3)^5 \equiv (-4)^5 \equiv (-5)^5 \pmod {11} \implies \boxed{a \equiv 1 \pmod {11}}. \] Case 4: ($b = 22$). Then by CRT we cannot have three consecutive fifth powers with the same remainder modulo $22$. (Take the congruences modulo $2$ and $11$). Hence the only solutions are those claimed at the beginning. $\blacksquare$

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