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

= 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 $$a^b = 2 c^d$$ 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):
 * 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
 * The second equation has infinitely many solutions no matter which b and d > 1 are taken, with the form $$v_p(a)=K d/GCD(b,d), v_p(a)=K b/GCD(b,d)$$ 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)