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

= 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 = (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.

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)