Wikipedia:Reference desk/Archives/Mathematics/2017 November 4

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

November 4

Two numbers with integer roots, n 2n

Are there two numbers n and 2n which both have integer roots? --Hofhof (talk) 13:24, 4 November 2017 (UTC)

Not integer square roots, because √2 is irrational. If you mean any integer roots, then √4 = ³√8 = 2. Double sharp (talk) 13:28, 4 November 2017 (UTC)

n=0 will do, but then n and 2n are not two numbers. Bo Jacoby (talk) 07:05, 5 November 2017 (UTC).

I confess I wasn't really thinking of 0. When I put on my number-theory hat, I don't think of 0 as a natural number; when I put on my set-theory hat, I do. Double sharp (talk) 07:26, 5 November 2017 (UTC)

Being even more pedantic every number n is a solution of n¹=n ;-) Dmcq (talk) 11:38, 5 November 2017 (UTC)

A multitasking number-theorist performs 2 or more tasks simultaneously. A multitasking set-theorist performs 0 or more tasks simultaneously. Bo Jacoby (talk) 12:16, 5 November 2017 (UTC).

• In the "any integer roots" interpretation of the question, the question is to find solutions of $a^{b}=2c^{d}$ with a,b,c and d integers. Assume the trivial cases away (i.e. a,c >0; b,d > 1).

Considering the p-adic order, we get $b*v_{2}(a)=1+d*v_{2}(c)$ and $b*v_{p}(a)=d*v_{p}(c)$ for prime p>2, this set of equalities being equivalent to the initial assertion. We can see that system as a set of equations on the independent variables $v_{p}(.)$ (this means the initial problem can be broken down into subproblems involving powers of prime numbers):

1. The first equation (power 2) has solutions if and only if GCD(b,d)=1, see Bezout's identity for proof and the description of those solutions
2. The second equation has infinitely many solutions no matter which b and d > 1 are taken, with the form $v_{p}(a)=Kd/GCD(b,d),v_{p}(a)=Kb/GCD(b,d)$ for K integer. (But per above, if a solution exists, GCD(b,d)=1)

So the answer to the general question is "infinitely many", with the powers being coprime in the general case. Tigraan^{Click here to contact me} 12:25, 5 November 2017 (UTC)

Nice general solution, thanks. Dmcq (talk) 11:46, 8 November 2017 (UTC)