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

Is this a joke? If a square has a prime factor ${\displaystyle p}$, it is divisible by ${\displaystyle p^{2}.}$If it has distinct prime factors ${\displaystyle p_{1},p_{2},...,p_{n}}$, it is divisible by the lcm of ${\displaystyle p_{1}^{2},p_{2}^{2},...,p_{n}^{2}}$, which, since all these are co-prime, equals ${\displaystyle p_{1}^{2}p_{2}^{2}\cdots p_{n}^{2}.}$So a square that is a multiple of ${\displaystyle i=p_{1}p_{2}}$is a multiple of ${\displaystyle i^{2}.}$ ‑‑Lambiam 21:35, 9 February 2025 (UTC)[reply]

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 = ${\displaystyle (2^{2})(3)}$. 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) = ${\displaystyle (2^{2})(3^{2})}$= 36, which is a perfect square that's smaller than ${\displaystyle i^{2}=12^{2}}$= 144.
• Second example: suppose i = 216. The prime factorization is 216 = 2x2x2x3x3x3 = ${\displaystyle (2^{3})(3^{3})}$. Notice that both 2 and 3 are raised to an odd power this time. Since you're looking for x such that (x)${\displaystyle (2^{3})(3^{3})}$is a perfect square, one way for that to occur is if you let x=(2)(3), in which case (x)(i) = ${\displaystyle (2^{4})(3^{4})}$, which is a perfect square that's smaller than ${\displaystyle i^{2}=((2^{3})(3^{3}))^{2}=(2^{6})(3^{6})}$. Another way for this to occur is if you let x = ${\displaystyle (2^{3})(3)}$, in which case (x)(i) = ${\displaystyle (2^{6})(3^{4})}$, which is again smaller than ${\displaystyle i^{2}=(2^{6})(3^{6})}$, and similarly if you let x = ${\displaystyle (2)(3^{3})}$.
• Third example: suppose i = 100. The prime factorization is 100 = 2x2x5x5 = ${\displaystyle (2^{2})(5^{2})}$. Notice that both 2 and 5 are raised to an even power this time. Since you're looking for x such that (x)${\displaystyle (2^{2})(5^{2})}$is a perfect square, one way for that to occur is if you let x=${\displaystyle 2^{2}}$, in which case (x)(i) = ${\displaystyle (2^{4})(5^{2})}$, which is a perfect square that's smaller than ${\displaystyle i^{2}=((2^{2})(5^{2}))^{2}=(2^{4})(5^{4})}$. Another way for this to occur is if you let x = ${\displaystyle (5^{2})}$, in which case (x)(i) = ${\displaystyle (2^{2})(5^{4})}$, which is again smaller than ${\displaystyle i^{2}=(2^{4})(5^{4})}$, and yet another way is if you let x = ${\displaystyle (3^{2})}$, in which case (x)(i) = ${\displaystyle (2^{2})(3^{2})(5^{2})}$, which is again smaller than ${\displaystyle i^{2}=(2^{4})(5^{4})}$.
• 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) = ${\displaystyle i^{2}}$, so there is no x that satisfies your condition of (x)(i) being smaller than ${\displaystyle i^{2}}$.

Hopefully you can now move to the general case, where the prime factorization is i = p₁^a₁p₂^a₂...p_n^a_n. If all of the ${\displaystyle a_{n}}$are equal to 1, there is no x that satisfies your condition. If one or more of the ${\displaystyle a_{n}}$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 = p₁^a₁p₂^a₂...p_n^a_n, and ${\displaystyle a_{1},a_{3}}$, and ${\displaystyle a_{4}}$are all of the odd powers, then the smallest x is x = ${\displaystyle (p_{1})(p_{3})(p_{4})}$, 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)[reply]

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

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