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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)[reply]

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)[reply]

February 10

Why this equation has no solutions ?

Take $SA$ , a semiprime, why is there no solution to this equation

(25² + x×SA)÷(67×y) == (z×SA)²

I’m meaning finding integers x and y and z such as the equation is true and where z≠0 ? 2A01:E0A:401:A7C0:E9E3:F9ED:9B83:6DB8 (talk) 14:36, 10 February 2025 (UTC)[reply]

I'm not certain what SA is supposed to be but if it is 1 and z is 1 then it is quite easy to get a solution of x=45 and y=10. NadVolum (talk) 15:24, 10 February 2025 (UTC)[reply]

Sorry, edited.

SA

is semiprime. 2A01:E0A:401:A7C0:C965:8D1D:CEAE:ABA0 (talk) 17:23, 10 February 2025 (UTC)[reply]

Multiplying both sides by

67y

and taking both sides modulo

SA

, we see that

25^{2}=5^{4}

would have to be congruent to

0\!\!{\pmod {SA}}

. In other words,

5^{4}

would have to be a multiple of a semiprime, which is impossible since it only has one prime factor. GalacticShoe (talk) 17:43, 10 February 2025 (UTC)[reply]

and if 25 was replaced by a square which is a multiple of the semiprime, would it works ? In reality 25 can be any perfect square, it doesn’t have to be 25. 2A01:E0A:401:A7C0:C965:8D1D:CEAE:ABA0 (talk) 02:59, 11 February 2025 (UTC)[reply]

The equation

{\frac {((m\times {\text{SA}})^{2})^{2}+x\times {\text{SA}}}{67\times y}}=(z\times {\text{SA}})^{2}

is a linear equation in

x.

It has a unique solution for any choice of non-zero

{\text{SA}},

m,

non-zero

y,

and

z.

If all of these are integers, then so is

x.

‑‑Lambiam 06:38, 11 February 2025 (UTC)[reply]

With

m

being a perfect square ? If yes, how to compute it ? 2A01:E0A:401:A7C0:C965:8D1D:CEAE:ABA0 (talk) 13:50, 11 February 2025 (UTC)[reply]

You wrote, "if 25 was replaced by a square which is a multiple of the semiprime". Well,

(m\times {\text{SA}})^{2}

is a square that is a multiple of

{\text{SA}}.

You don't need to compute

m

;

m

can be set to any value that you wish, including squares, so you can take

m=100,

or

m=1,

or even

m=0.

Aren't you overdoing the squaring? If

m=a^{2}

and you expand the brackets, you end up with

{{a^{2}}^{2}}^{2}{{\text{SA}}^{2}}^{2}=a^{8}{\text{SA}}^{4}.

‑‑Lambiam 21:30, 11 February 2025 (UTC)[reply]

Yes, that’s correct, with the ²

m

is a perfect square by construction. But how to find a value of

m

such as a solution of

x

and

y

and

z

exists ? 2A01:E0A:401:A7C0:88B1:8A9D:F4E0:D37A (talk) 07:24, 13 February 2025 (UTC)[reply]

Take

m=1.

‑‑Lambiam 09:27, 13 February 2025 (UTC)[reply]

February 14

Could eversion be a functor?

After reading Sphere eversion, which I didn't understand(not the article's fault), I wonder if there is a class of objects including the 3 particular spheres, and maybe tori that can be everted and form a category, in which eversion would be a functor from that category to itself.Rich (talk) 16:16, 14 February 2025 (UTC)[reply]

Presumably, in a categorical treatment, the objects would be self-intersection-free immersions of bounded and connected 2-dimensional smooth orientable manifolds in three-dimensional space. The morphisms would be, presumably, regular homotopies. Some of these morphisms preserve orientation; the others are "disorienting" – they flip the orientation around. Some morphisms are endomorphisms: their domain and codomain are the same object. Eversions are then the disorienting endomorphisms, a subclass of the morphisms. This subclass is not particularly interesting, categorically speaking, since the composition of two eversions is not an eversion. We can form a subcategory whose objects are the evertible objects, but we cannot restrict the morphisms to just the eversions, because the identity morphisms would thereby be excluded.

An endofunctor of this category should map morphisms to morphisms; eversions, being themselves morphisms, live at a lower categorical level. I see no obvious way of defining another notion of morphism between evertible objects in which an endofunctor might correspond to the notion of eversion. ‑‑Lambiam 06:05, 15 February 2025 (UTC)[reply]

February 16