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March 16

Smallest triangular number with prime signature the same as A025487(n)

Smallest triangular number with prime signature the same as (sequence A025487 in the OEIS)(n), or 0 if no such number exists. (there is a similar sequence (sequence A081978 in the OEIS) in OEIS)

For n = 1 through n = 29, this sequence is: (I have not confirmed the n = 15 term is 0, but it seems that it is 0, i.e. it seems that there is no triangular number of the form p^3*q^2 with p, q both primes)

1, 3, 0, 6, 0, 28, 0, 136, 66, 0, 36, 496, 276, 0, 0, 118341, 120, 0, 1631432881, 300, 8128, 210, 0, 528, 0, 29403, 1176, 32896, 630

Is it possible to extend this sequence to n = 100? 1.165.194.85 (talk) 00:51, 16 March 2024 (UTC)

This paper by Cohn implies that there are no prime solutions $(p,q)$ to $p^{3}+1=2q^{2}$ and $2p^{2}+1=q^{3}$. If anyone here can prove that the elliptic curves $y^{2}=x^{3}\pm 4$ have no nontrivial integral points (for $-4$ that would be $(2,\pm 2)$ and $(5,\pm 11)$, while for $4$ that would be $(0,\pm 2)$), then that implies that there are no triangular numbers of the form $p^{3}q^{2}$ at all. GalacticShoe (talk) 06:22, 16 March 2024 (UTC)

I'm not sure what you mean by "at all". If you drop the requirement that q is prime then there are many solutions, for example 242⋅243/2 = 3³33², 12167⋅12168/2 = 23³78². --RDBury (talk) 08:42, 16 March 2024 (UTC)

Sorry, shoulda clarified; no triangular numbers of the form $p^{3}q^{2}$ at all where $p,q$ are prime. Naturally, the two cases mentioned earlier (the latter example of which I hastily posted before remembering that I completely forgot about the prime signature part) are either not covered by Cohn's paper, with the $242$ case being invalid since we spread powers of $3$ among the cube and the square, or they are one of Cohn's special case, as with $12167$. GalacticShoe (talk) 09:51, 16 March 2024 (UTC)

Thanks for the clarification. I don't have access to the Cohn paper, but it seems to me that the p³+1=2q² is trivial to exclude, and similarly for p³-1=2q². That leaves 2p³±1=q². Using brute force I found 2⋅23³+2=156² but none, even when p is only required to be odd and q can be anything, where the difference is one. Note that this generates the 12167 example, but that's not the way I found it. I'm not sure how y²=x³±4 is related, but I'm no expert on elliptic curves so I may have to take your word on that. In any case, given that the p³q² is proving so difficult, it seems unlikely that getting to n=100 is feasible. Finding entries where there is a solution should only require a computer search, but if none are found then proving that there are none to be found can be difficult. It seems ironic that while there is apparently no p³q² solution, there is a p⁵q², though this seems less likely at first glance. RDBury (talk) 18:24, 16 March 2024 (UTC)

The $y^{2}=x^{3}\pm 4$ condition arises because, if we write $x=2p$ and $y=2q$, then an integer solution to $y^{2}=x^{3}\pm 4$ implies $(2q)^{2}=(2p)^{3}\pm 4\Rightarrow 4q^{2}=8p^{3}\pm 4\Rightarrow q^{2}=2p^{3}\pm 1$ is a rational, possibly integer solution to the equalities we are looking at. All integer solutions to $q^{2}=2p^{3}\pm 1$ can be generated this way, so the lack of a nontrivial integer solution to $y^{2}=x^{3}\pm 4$ (or the existence only of solutions where $x,y$ are odd) would rule out any further triangular numbers of the form $p^{2}q^{3}$ with $p,q$ prime. GalacticShoe (talk) 21:33, 16 March 2024 (UTC)

Right, I should have seen that. The 2p³+1=q² case may be easier than I thought; it reduces to 2p³=(q+1)(q-1). You can rule out p=2, q=2, giving q odd, but then by unique factorization q±1 divides p, which is impossible. I thought of applying a similar trick to 2p³-1=q², or 2p³=(q+i)(q-i), and using unique factorization over Gaussian integers. But I quickly got bogged down with this. --RDBury (talk) 02:10, 17 March 2024 (UTC)

A similar disproof is that after proving $p=2$ and $q=2$ impossible by simple casework, $q$ odd implies that $(q+1)(q-1)$ is a multiple of $4$, which $2p^{3}$ cannot be. I imagine that $2p^{3}-1=q^{2}$ is indeed not as simple to pin down. GalacticShoe (talk) 02:45, 17 March 2024 (UTC)

So if $y^{2}=x^{3}\pm 4$ has no integer solution other than $(0,\pm 2)$, $(2,\pm 2)$, $(5,\pm 11)$, than a(15) in the above sequence is 0? How about a(30), a(31), a(32), …? Do you have the status for a(n) for all $n\leq 100$? (sequence A081978 in the OEIS) has the proof for which some prime signatures do not exists for triangular numbers such as $p^{6}q^{2}$ (i.e. a(33) = 0), see the comment of Jinyuan Wang in Aug 22 2020. (You can also check whether a(1) ~ a(29) which I have listed above are correct) (a(n) is the smallest triangular number with prime signature the same as (sequence A025487 in the OEIS)(n), or a(n) is 0 if no such number exists) 61.224.130.152 (talk) 03:21, 18 March 2024 (UTC)

I mentioned this a few paragraphs back. Given that finding the value of a single entry is turning out to be a project in itself, and that it seems unlikely that computations will get any easier going on, I don't see getting to 100 as feasible. A081978 is asking for less information, and even so it took some high powered number theory to get it's values. --RDBury (talk) 15:36, 18 March 2024 (UTC)

March 18

Tautology logical reasoning

Does every tautology have a verbal logical reasoning not requiring any use of truth tables or logical connectives, just like combinatorial identities/formulae (at least many) have combinatorial reasonings?
For example, this tautology here: ${\bigl (}a\to (b\to c){\bigr )}\to {\bigl (}b\to (a\to c){\bigr )}$. יהודה שמחה ולדמן (talk) 18:44, 18 March 2024 (UTC)

Well one can always read out the maths! But it would become very tedious and you might lose attention for anything much more complicated than the above. That though can be transformed by saying: A implies B means A is false or B is true. Then A implies B implies C means the same as A is false or B is false or C is true and that is the same as B is false or A is false or C is true. I mean what are you looking for beyond that? To start using the ancient business of syllogisms with names like Baroco? NadVolum (talk) 19:30, 18 March 2024 (UTC)

I mean literally, how do we deduce with logical reasoning that the initial part $a\to (b\to c)$ implies the final part $b\to (a\to c)$?

Of course I understand that we use symbols to express arguments, such as $a\to b=(\neg a\land \neg b)\oplus (\neg a\land b)\oplus (a\land b)=\neg a\lor b$.

But I am also asking if this is a definition apriori (${\stackrel {\rm {def}}{=}},{\stackrel {\triangle }{=}}$) or a theorem. יהודה שמחה ולדמן (talk) 20:53, 18 March 2024 (UTC)

If I tell you that $(a\,\natural \,(b\,\natural \,c))\,\natural \,(b\,\natural \,(a\,\natural \,c))$ is a tautology without supplying an interpretation of the connective "$\natural$" and ask you to provide verbal reasoning for this tautology, you will not be able to make any inroads. You can't do anything if you don't know the meaning of "$\natural$". So it is not clear to me what kind of "verbal reasoning" might correspond to this tautology without referencing the interpretation of the material conditional connective "$\rightarrow$". To me, this would be like requiring a verbal proof of $a+b=b+a$ without any use of addition. The tautology can be verbalized: "If we know that if A is true, then if B is true then C is true, it follows that if B is true, then if A is true then C is true." If you can wrap your head around it, this should be self-evident, but a proof by natural deduction of the tautology can also be verbalized. ("Assume we know that if A is true, then if B is true then C is true. We want to show that it follows that if B is true, then if A is true then C is true. So assume that B is true. ...") However, unless one has an uncanny ability, one will get lost in keeping track of which assumptions hold at various steps and which have been discharged.  --Lambiam 08:49, 19 March 2024 (UTC)

March 20

Whether it coincides with a simpler function

Is y = sin (arcsin x) (1) the same function as y=x, if we consider all branches of logarithm (of any real number) and all branches of inverse sine function? Or does (1) remain meaningless for any argument outside the range [-1;1] when we restrict it to real value for both the domain and the image, and (1) will coincide with the identity function only when we regard it as a function that map complex numbers to complex number? Does the logarithm of negative numbers lead to the presence of removable singularities for (1)? (In contrast, the function y=x obviously does not contain any singularity). I was able to prove that y = arcsin (sin x) and y = sin (arcsin x) are not always the same, but I still can't settle the aforementioned problems. 2402:800:63AD:81DB:105D:F4F:3B26:74C5 (talk) 14:36, 20 March 2024 (UTC)

A univalued function and a multivalued function $f:X\to Y,$ possibly partial, can be represented by a relation $R_{f}\subseteq X\times Y.$ The total identity function ${\text{id}}:X\to X$ corresponds to the identity relation $I_{X}=\{(x,x)|x\in X\}\subseteq X\times X.$ Function composition corresponds to relation composition: $R_{(g\,\circ f)}=R_{f}\,;R_{g}.$ The multivalued function inverse correspond to relation converse: $R_{(f^{{-}1})}=(R_{f})^{\text{T}}\!.$

Just like the multivalued complex logarithm is the multivalued inverse of the exponential function $\exp :\mathbb {C} \to \mathbb {C}$, the complex $\arcsin$ including all branches is the multivalued inverse of function $\sin :\mathbb {C} \to \mathbb {C} .$ So

$R_{(\sin \circ \arcsin )}=R_{\arcsin }\,;R_{\sin }=(R_{\sin })^{\text{T}};R_{\sin }.$

Generalizing this from the sine function to an arbitrary (univalued) function $f$, we have:

$(x,y)\in (R_{f})^{\text{T}};R_{f}\Leftrightarrow \exists z:((z,x)\in R_{f})\land (z,y)\in R_{f})\Leftrightarrow \exists z:(x=f(z)\land y=f(z)).$

Clearly, this implies $x=y,$ so the composed relation is the identity relation on the range of $f,$ representing the identity function on that range.  --Lambiam 18:21, 20 March 2024 (UTC)

March 22

Sin, cos and ellipses

In this book [1] (linked to the right page), left column towards the bottom. I'm having a problem with the "hence."

I understand that:

A: the point M has the coordinates (x,y), which is also (sin φ, cos φ), no matter how you slide the straight KL, for all φ.

B: the formula for the ellipse. $${\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1.$$

I just don't get how B follows from A. Maybe I'm missing a concept that the authors take for granted. But shouldn't the text have explained something in between? Something like why the ellipse matches the Pythagorean trigonometric identity? Or why the set of all possibles values of M is $\sin ^{2}\theta +\cos ^{2}\theta =1$? Why is that when you have the coordinates (sin, cos) you add them and equal to 1 to produce an ellipse? Grapesofmath (talk) 17:27, 22 March 2024 (UTC)

I think you've misread the text: the point M (x, y) is actually (a sin φ, b cos φ). So x/a = sin φ and y/b = cos φ. Substituting into the identity $\sin ^{2}\phi +\cos ^{2}\phi =1$ (which is true for all φ) gives the formula in (B). AndrewWTaylor (talk) 18:09, 22 March 2024 (UTC)

More precisely, the text identifies the point M (x; y) by "y = b sin φ and x = a cos φ". The resulting equation is the same.  --Lambiam 19:32, 22 March 2024 (UTC)

Thanks for the feedback. But shouldn't the text be explicit here and explain that A applied to Pythagorean trigonometric identity result in B? Isn't this a jump too big in the train of thought? Grapesofmath (talk) 23:59, 22 March 2024 (UTC)

It's not immediately obvious, but it's not that hard either given the diagram. The y = b sin φ come from the lower right triangle and x = a cos φ comes from the upper left triangle. Once you have those equations, the equation (1) in the text follows as explained above. I think as a reader you're meant to figure out this kind of detail yourself. The alternative would be that the text becomes long-winded and pedantic. It's also a better learning experience for the reader if they have to think about the text as they reading and fill in some missing steps. A lot depends on the intended audience as well; apparently the book in meant for people with a certain basic knowledge of geometry, perhaps with some experience writing proofs. (That kind of information is often given in an introduction, but this is the introduction.) --RDBury (talk) 04:17, 23 March 2024 (UTC)