Testwiki:Reference desk/Archives/Mathematics/2025 February 9

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February 9

How to find a multiplier so that the resulting number is a small perfect square…

Simple question, let’s take a semiprime $i$.
I want to find a number $x$ such as x×i is perfect square (the square root is an integer) but the resulting square is small than $i^2$2A01:E0A:401:A7C0:9407:59CB:EFF:777C (talk) 20:36, 9 February 2025 (UTC)

Is this a joke? If a square has a prime factor $p$, it is divisible by $p^2.$ If it has distinct prime factors $p_1,p_2,...,p_n$, it is divisible by the lcm of $p_1^2,p_2^2,...,p_n^2$, which, since all these are co-prime, equals $p_1^2{p}_2^2\cdots p_n^2.$ So a square that is a multiple of $i=p_1{p}_2$ is a multiple of $i^2.$  ‑‑Lambiam 21:35, 9 February 2025 (UTC)

IP, I don't know if you're still looking for an answer, but this is only possible for some numbers i. I'll start with a few examples before trying to discuss the general case.

• First example: suppose i = 12. The prime factorization is 12 = 2x2x3 = $(2^2)(3)$. Notice that 2 is raised to an even power, but 3 is raised to an odd power. Since you're looking for x such that (x)(2^2)(3) is a perfect square, each prime in the the prime factorization of (x)(i) must be raised to an even power. In this case, if x=3, then (x)(i) = $(2^2)(3^2)$ = 36, which is a perfect square that's smaller than $i^2=12^2$ = 144.
• Second example: suppose i = 216. The prime factorization is 216 = 2x2x2x3x3x3 = $(2^3)(3^3)$. Notice that both 2 and 3 are raised to an odd power this time. Since you're looking for x such that (x)$(2^3)(3^3)$ is a perfect square, one way for that to occur is if you let x=(2)(3), in which case (x)(i) = $(2^4)(3^4)$, which is a perfect square that's smaller than $i^2=((2^3)(3^3){)}^2=(2^6)(3^6)$. Another way for this to occur is if you let x = $(2^3)(3)$, in which case (x)(i) = $(2^6)(3^4)$, which is again smaller than $i^2=(2^6)(3^6)$, and similarly if you let x = $(2)(3^3)$.
• Third example: suppose i = 100. The prime factorization is 100 = 2x2x5x5 = $(2^2)(5^2)$. Notice that both 2 and 5 are raised to an even power this time. Since you're looking for x such that (x)$(2^2)(5^2)$ is a perfect square, one way for that to occur is if you let x=$2^2$, in which case (x)(i) = $(2^4)(5^2)$, which is a perfect square that's smaller than $i^2=((2^2)(5^2){)}^2=(2^4)(5^4)$. Another way for this to occur is if you let x = $(5^2)$, in which case (x)(i) = $(2^2)(5^4)$, which is again smaller than $i^2=(2^4)(5^4)$, and yet another way is if you let x = $(3^2)$, in which case (x)(i) = $(2^2)(3^2)(5^2)$, which is again smaller than $i^2=(2^4)(5^4)$.
• Fourth example: suppose i = 30. The prime factorization is 30 = 2x3x5, where each prime is raised to an odd number, and all of the powers are equal to 1. The problem in this example is that the smallest x such that (x)(i) is a perfect square is x = (2)(3)(5). But in this case (x)(i) = $i^2$, so there is no x that satisfies your condition of (x)(i) being smaller than $i^2$.

Hopefully you can now move to the general case, where the prime factorization is i = Template:Math. If all of the $a_n$ are equal to 1, there is no x that satisfies your condition. If one or more of the $a_n$ are greater than 1, then you choose x such that all of the primes in the the prime factorization of (x)(i) will be raised to an even power. The smallest x will be the product of the primes in the prime factorization of i that are raised to an odd power (e.g., if i = Template:Math, and $a_1,a_3$, and $a_4$ are all of the odd powers, then the smallest x is x = $(p_1)(p_3)(p_4)$, and you'll also be able to find some other values for x by multiplying that smallest value by a small perfect square). FactOrOpinion (talk) 20:59, 16 February 2025 (UTC)

Note, though, that this implies that $i$ is not a semiprime as supposed in the question.  ‑‑Lambiam 04:14, 17 February 2025 (UTC)

You're right, I missed that $i$ is a semiprime. I also missed the trivial solution $x$ = 0. If we add the constraint that $x$ must be positive, then $x$ exists iff $i$ = $p^2$ for a prime $p$ > 2. Just let $x$ = $n^2$ for any whole number $n$ with 0 < $n$ < $p$. FactOrOpinion (talk) 03:35, 18 February 2025 (UTC)