Wikipedia:Reference desk/Archives/Mathematics/2023 May 15

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May 15

Can 2*12^n+1 be square for n>2?

2*12^n+1 is square for n=1 and n=2, but can 2*12^n+1 be square again for n>2? 210.244.74.74 (talk) 21:04, 15 May 2023 (UTC)

It can't. Assume otherwise. Expand out the expression as $2^{2n+1}3^{n}+1=k^{2}$, so that $2^{2n+1}3^{n}=(k-1)(k+1)$. Since $(k+1)-(k-1)=2$, and $n>2$, $3^{n}$ divides one term entirely.

If $3^{n}|k-1=2^{a}3^{n}$ for $0\leq a\leq 2n+1$, then $2^{a}3^{n}+2=2^{2n-a+1}$. $a=0$ would imply that $3^{n}+2=2^{2n-1}$, which for $n>2$ can't be true by parity. So $a>0$ (and also by evenness, $a<2n+1$.) Divide both sides by $2$ to get $2^{a-1}3^{n}+1=2^{2n-a}$. $a=2n$ would once again lead to a problem, as $2^{2n-1}3^{n}=0$ would not make sense, so $2^{2n-a}$ is even, implying that $2^{a-1}3^{n}$ is odd, and thus $a=1$. So $3^{n}+1=2^{2n-1}$. There is no integer $n>2$ for which this is the case, so we discard this case.

Now if $3^{n}|k+1=2^{a}3^{n}$ for $0\leq a\leq 2n+1$, then $2^{a}3^{n}=2^{2n-a+1}+2$. $a=0$ would lead to $3^{n}=2^{2n+1}+2$, which would be impossible by parity, so $a>0$. If $a=2n+1$, then $2^{2n+1}3^{n}=3$, which cannot be for $n>2$, so $a<2n+1$. Thus we can divide both sides by $2$ to get $2^{a-1}3^{n}=2^{2n-a}+1$. $a=2n$ once again would again lead to the impossible $2^{2n-1}3^{n}=2$, so the right side is odd. Thus, $a=1$ and $3^{n}=2^{2n-1}+1$. This time, there are integer solutions to this, but they are $n=1,2$. So we discard this case.

But since those are the only possible cases, there's a contradiction; so no $n>2$ has the property that $2*12^{n}+1$ is square. GalacticShoe (talk) 23:28, 15 May 2023 (UTC)

A problem of the Aliquot sum

Let s(n) = sigma(n)-n = OEIS: A001065(n), s^k is the iterated function, we list the largest k such that a given natural number n is in the range of s^k.

k	such natural numbers n	OEIS sequence
0	2, 5, 52, 88, 96, 120, 124, 146, 162, 188, 206, 210, 216, 238, 246, 248, 262, 268, 276, 288, 290, 292, 304, 306, 322, 324, 326, 336, 342, 372, …	OEIS: A005114
1	208, 250, 362, 396, 412, 428, 438, 452, 478, 486, 494, 508, 672, 712, 716, 772, 844, 900, 906, 950, 1042, 1048, 1086, 1090, 1112, 1132, 1140, 1252, 1262, 1310, 1338, 1372, …	OEIS: A283152
2	388, 606, 696, 790, 918, 1264, 1330, 1344, 1350, 1468, 1480, 1496, 1634, 1688, 1800, 1938, 1966, 2006, 2026, 2202, 2220, 2318, 2402, 2456, 2538, 2780, 2830, 2916, 2962, 2966, 2998, …	OEIS: A284147

Can you find the sequence of k=infinity (i.e. n is in the range of s^k for all natural numbers k)? Assuming the strong version of Goldbach conjecture is true, i.e. all even number >6 are the sum of two distinct primes. 210.244.74.74 (talk) 21:15, 15 May 2023 (UTC)