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

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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)