PDatK40SP wrote: We can also prove that $m \equiv 2 \ (mod \ 12)$ ( and replace $m$ by $12t+2$ in the formula of $A$, we will have $A$ is odd immediately ). Use this lemma: ( I think it's well-known ) Let $n$ is an odd integer, and $k$ is an odd divisor of $3^{n}+1$. Then, $k \equiv 1 \ (mod \ 6)$ $m|(m+3)^{n}+1 \Rightarrow m|3^{n}+1$ Of course $3 \not | m$, so if $n$ is even, then $(m+3)^{n}+1 \equiv 2 \ (mod \ 3)$, which is impossible because $3m|(m+3)^{n}+1$. So $n$ is even, then, from the lemma, we have $m = 2^{k}(6t+1) \ \ (k,t \in \mathbb{N})$ If $k$ is even, we have $(m+3)^{n}+1 \equiv 2 \ (mod \ 3)$, which is impossible. Moreovers, easy to see that $8 \not | 3^{n}+1 \ \forall n \in \mathbb{N}$ so $k \geq 3$ is impossible. So, $k=1 \Rightarrow m=12t+2$

Peter wrote: PDatK40SP wrote: Let $ n$ is an odd integer, and $ k$ is an odd divisor of $ 3^{n}+1$. Then, $ k\equiv 1\ (mod\ 6)$ Why?

Студент wrote: Peter wrote: PDatK40SP wrote: Let $ n$ is an odd integer, and $ k$ is an odd divisor of $ 3^{n}+1$. Then, $ k\equiv 1\ (mod\ 6)$ Why? It's equivalent to showing that none of its prime divisors can be 5 modulo 6. Let $ a = 3^{(n+1)/2}$ and suppose $ 3^{n}+1$ is divisible by a prime $ p\equiv 5\pmod{6}$. Then $ a^{2}\equiv-3\pmod{p}$. At this point, Quadratic Reciprocity would work. You could also show that \[ x=2^{-1}(a-1)\Rightarrow x^{2}+x+1\equiv 0\pmod{p}\] So $ x^{3}\equiv 1\pmod{p}$ and $ x^{p-1}\equiv 1\pmod{p}$. We have $ p-1$ and 3 relatively prime, so $ x\equiv 1\pmod{p}$. But as $ x^{2}+x+1\equiv 0\not\equiv 3\pmod{p}$, this is a contradiction.

More : 1) There exist infinite $ (m,n)\in N$ satisfy the condition. 2)$ m\equiv 2(\mod 12)$ (take from my teacher).

Peter wrote: PDatK40SP wrote: Use this lemma: ( I think it's well-known ) Let $ n$ is an odd integer, and $ k$ is an odd divisor of $ 3^{n} + 1$. Then, $ k \equiv 1 \ (mod \ 6)$ what's well known lemma?

i will fakesolve ur problem

TTsphn wrote: More : 1) There exist infinite $ (m,n)\in N$ satisfy the condition. 2)$ m\equiv 2(\mod 12)$ (take from my teacher). I agree with your Teacher. Except of some ifs and buts. According to the mysterious Lemma the solution of the problem A17 is: m = 2 + 12t for ANY natural t, including t = 0. Those t's may also be perfect squares like t = 4, 9, 16, 25,... so that m = 50, 110, 194, 302,.. All those m's make A an odd number, at least for certain n's. Right? If you agree, let me offer you my CONJECTURE A17. : If t > 1 is a perfect square, then A is not an integer, for ANY n. Unable to prove or disprove my conjecture. If I'm right, then maybe Peter's Lemma has to be modified.(?) Please help. Thanks.

Peter, The first sentence of your post. You “…will have A odd immediately”(?) But what if t = 4, 9, 16, 25, 36 , 49…or 12, 5, 19 and n is any number. Any! Then A will be proper or improper fraction - not an odd as you claim. If t = 0 AND n = 1, 3, 5, 7,... or t = 7 AND n = 21, 63, 105, 147, …. or t = 22 AND n = 9, 27, 45, 63, 81…. then in each case we get infinite number of odd A’s “immediately” and "m = 2 + 12t" works. Nothing is always You ignore(?) that A is not A(t), but A(t, n). t is parameter, n is a variable (see below) Trivial: A can be either a fraction (for any n) or an odd (for certain t’s and certain unlimited n’s). Let’s subdivide the set of 26 consecutive t’s (from 0 to 25) by two subsets: ‘bad’ t’s and ‘good’ t’s. Proposition 1. ‘Bad’ t’s do not make A an odd. For any n. Here they are: t = 4, 9, 12, 14, 15, 16, 18, 19, 20, 21, 23, 24, 25. Proposition 2. ‘Good’ t’s do make A an odd. But not for any n. Good t’s: t = 0, 1, 2, 3, 5, 6, 7, 8, 10, 11, 13, 17, 22 Bad t’s do not solve the problem A17. We’ll ignore them from now on. Now we’ll find all unlimited odd A’s (all of them) for first 13 good t’s and and predefined n’s We ‘ll show that A can not be just any odd but unlimited strictly defined ones. In the table below k runs through all odds: k = 1, 3, 5, 7, …. In A(t, n): t is parameter, n is variable (a linear function of odd k’s for each ‘good’ t). Some examples: t = 0 A(0, k) is an odd for any n = k = 1, 3, 5, 7, ……. then A(0, k) = 1, 21, 521, 13 021, 325 521, … t = 1 A(1, 3k) is an odd for any n = 3k = 3, 9, 15, 21….. then A(1, 3k) = 117, 2 823 520 869, ……… t = 7 A(7, 21k) is an odd for any n = 21k = 21, 63, 105, 147,… then A(7,21k) =…..infinite number of huge odds t = 13 A(13, 39k) is an odd for any n = 39k = 39, 117, 195,… then A(13, 39k) = ….. ‘’ ‘good’ t___m=2+12t_____n_______A(t,n) = ((5+12t)^n + 1)/3(2+12t)_____A(t, n), all odds. ============ ================================================================== 0_________2__________k________A (0, k) = (5^k + 1)/6_____________A(0, k) = 1, 21, 521, 13021, 325521,. 1________14_________ 3k_______A (1, 3k) = (17^3k + 1)/42_________ A(1, 3k) = 117, 2823520869, ….. 2________26__________3k_______A(2, 3k) = (29^3k + 1)/78_________ A(2, 3k) = Infinite number of huge odds 3________38__________9k_______A(3, 9k) = (41^9k + 1)/114________ A(3, 9k) = _______ same 5________62_________15k_______A(5,15k) = (65^15k + 1)/186_______A(5, 15k)________ same 6________74__________9k_______A (6, 9k) = (77^9k + 1)/222_______ A(6, 9k)_________ same 7________86_________ 21k______ A(7, 21k) = (89^21k + 1)/258_____ A(7, 21k)________ same 8________98__________21k______A(8, 21k) = (101^21k + 1)/294_____A(8, 21k)________ same 10______122___________5k______A(10, 5k) = (125^5k + 1)/366______ A(10, 5k)__________" 11______134__________11k______A(11, 11k) = (137^11k + 1)/402_____A(11, 11k)__________" 13______158__________39k______A(13, 39k) = (161^39k + 1)/474_____A(13, 39k)_________‘’ 17______206__________17k______A(17, 17k) = (209^17k + 1)/618_____A(17, 17k)_________‘’ 22______266___________9k_____ A(22, 9k ) = (269^9k + 1)/798______A(22, 9k)___________‘’ There are 13 good and 13 bad t's. Just coincidence. That’s all I can say about that. As an exercise: for 25 < t < 51 find all ‘good’ t’s and corresponding n’s.

all odd prime factors of m are $\equiv 1 \pmod{3}$ write $m=2^k \cdot j$, $k \geq 0$ and $j\equiv 1 \pmod{3}$ taking modulo 3, $m\equiv 2\pmod{3}$ and n is odd $2^k \cdot j \equiv 2 (mod 3) \implies$ k is odd but $k\geq 3 \iff 8|m \implies 8|3^n+1$ creates contradiction so $k=1 \iff 2||m$ so $m\equiv 2\pmod {12}$ $(m+3)^n+1\equiv 2\pmod4$ , so A is odd one example is $(m,n,A)=(2,1,1)$

Peter wrote: Let $m$ and $n$ be natural numbers such that \[A=\frac{(m+3)^{n}+1}{3m}\]is an integer. Prove that $A$ is odd. When $m$ is odd ,$ A$ is always odd because both numerator and denominator are odd. Claim: When $ m,n$ are both even ,or when $ m$ is divisible by $3$ then $ A$ is never an integer. Proof If $m \equiv 0 \mod {3}$ then $$(m+3)^n+1 \equiv m^n+1 \equiv 1 \mod 3 $$wich contradicts the fact that $ A$ is integer . When $m \not\equiv 0 \mod 3$ and $n $ even we have that$$(m+3)^{2k}+1 \equiv m^{2k}+1 \equiv 2 \mod 3 $$wich again contradicts that $A$ is integer Claim: When $m$ is even , then we need $m=2k $ or $m= 4k$ where $k$ is odd integer for $A $ to be integer. Proof: We must have $$ (m+3)^n+1 \equiv 3^n+1\equiv 0 \mod m$$or $$ m| 3^n+1$$since $n$ is odd By LTE we can say that $$v_2(m) \leq v_2 (3^n+1)=v_2(4)+v_2(n)=2 $$ Claim: When $m=2k$ or $m=4k$ then $v_2(A)=0$ Proof $$v_2(A)=v_2 ((m+3)^n+1)-v_2(3m)=v_2(2k+4)-v_2(6k)=0$$for $ m=2k$ and $$v_2(A)=v_2(4k+4)-v_2(12k)=0$$for $m=4k$ After checking all cases we can say that $A$ is always odd. $\blacksquare$

XbenX wrote: Claim: When $m=2k$ or $m=4k$ then $v_2(A)=0$ Proof $$v_2(A)=v_2 ((m+3)^n+1)-v_2(3m)=v_2(2k+4)-v_2(6k)=0$$for $ m=2k$ and $$v_2(A)=v_2(4k+4)-v_2(12k)=0$$for $m=4k$ This part is wrong. Suppose we have $m \equiv 4 \text{ (mod }8 \text{)}$ and $n$ an odd integer, then $(m+3)^n$ is congruent to $7$ (mod $8$). then $v_2(A)$ might be positive, and in that case we need to prove the divisibility relation can't hold, instead of computing $v_2(A)$ more.

Legandre symbol

Note that by mod $3$ we have $m\equiv 2\pmod 3$ and $n$ is odd. Now we'll prove a lemma (sadly, I couldn't come up with the idea of the proof for this myself ) Lemma: If $n$ is odd and prime $p\mid 3^n+1$ then $p\equiv 1 \pmod 3$. Proof. Note that since $n+1$ is even and $-3\equiv 3^{n+1}\pmod p$, it follows that $-3$ is a quadratic residue modulo $p$. So $\left(\frac{3}p\right)=(-1)^{\frac{p-1}2}$. Now by quadratic reciprocity we have $$\left(\frac{3}p\right)\left(\frac p3 \right)=(-1)^{\frac{p-1}2\frac{3-1}2}=(-1)^{\frac{p-1}2}\Rightarrow \left(\frac p3 \right)=1.$$Thus $p$ is a quadratic residue modulo $3$. It follows that $p\equiv 1 \pmod 3$, as required. //// Suppose that $v_2\left ((m+3)^n+1\right)\geq 3$, then by mod $8$ analysis $m\equiv 4 \pmod 8$ and so $v_2(m)=2$. Suppose $m=2^2 \prod p_i^{e_i}$. Then $p_i\mid 3^n+1$, so it's $1$ mod $3$ by the lemma. But then $m\equiv 1\pmod 3$, a contradiction. So, $v_2\left ((m+3)^n+1\right)\leq 2$ and the rest is easy (too exhausted to write...)