A specific Diophantine equation related to the congruent number question

Let $$n$$ be an odd square free natural number. J.B. Tunnel in his 1983 paper, showed that a number $$n$$ is congruent, if and only if the number of triples of integers satisfying $$2x^2+y^2+8z^2=n$$ is equal to twice the number of triples of integers satisfying $$2x^2+y^2+32z^2=n$$. This is by assuming the BSD conjecture, but still we do not know an efficient (polynomial time) algorithm to determine whether a number is congruent or not, from the theorem stated above.

I am trying to move with this problem a little. A simple observation tells us that, if $$(alpha,beta,gamma)$$ satisfies $$2x^2+y^2+32z^2=n$$, then $$(alpha,beta,2gamma)$$ and $$(alpha,beta,-2gamma)$$ satisfies $$2x^2+y^2+8z^2=n$$. So, like this we deduce that if $$2x^2+y^2+8z^2=n$$ has twice integral solution than $$2x^2+y^2+32z^2=n$$, then $$2x^2+y^2+8z^2=n$$ cannot have its integral solution with $$z$$ odd.

So, now the problem reduces to for what values of $$n$$, we will have an integral solution of $$2x^2+y^2+8(2z+1)^2=n$$. If it has a solution then $$n$$ is not congruent, otherwise it is. Now, I do not know how to proceed any further. As, the equation is not homogenous, one cannot directly invoke Hasse Minkowski’s local global principle, so trying to solve over $$p$$-adics is not an option. However, if one fails to find solution over $$mathbb{Q}_p$$ for any $$p$$ for a particular $$n$$, then $$n$$ is congruent. By this, I was able to prove that the numbers $$nequiv 5$$ mod $$8$$ and $$nequiv 7$$ mod $$8$$ are always congruent, as this type of numbers were failing to give any solution mod $$8$$ and hence, there was no $$mathbb{Q}_2$$ solutions. But this will never say whether a number is not congruent.

I do not have any idea to proceed with the problem. Again, the diophantine problem is for what $$n$$, does $$2x^2+y^2+8(2z+1)^2=n$$ has integral solutions. Any suggestions or directions to move will be really helpful.

nt.number theory – Semi generalized solution to many congruent number cases

All congruent number problem square free integer N values can we written as equal or smaller square free integers $$N=n_1*n_2*n_3*n_4$$

And always have at least one $$n_1$$,$$n_2$$,$$n_3$$,$$n_4$$ set of values that satisfy the following two equations for integers $$d_1$$,$$d_2$$,$$d_3$$,$$d_4$$

$$n_1 d_1^2 + n_2 d_2^2 = n_3 d_3^2$$
and
$$n_1 d_1^2 + 2 n_2 d_2^2 = n_4 d_4^2$$

Where rational t,x,y,z

$$t=n_1 d_1^2 / (n_2 d_2^2)$$

x=2+t
y=2+2/t
z=2+t+2/t

xy/2=N(squarefactor)^2

The key thing here is that if you randomly generate congruent numbers by randomly picking $$n_1$$,$$n_2$$,$$d_1$$,$$d_2$$ and calculating $$n_3$$,$$d_3$$ and $$n_4$$,$$d_4$$ you will find that N is nearly always greater than $$d_1$$,$$d_2$$,$$d_3$$ or $$d_4$$. So all $$d_1$$,$$d_2$$,$$d_3$$,$$d_4$$ values can be found by guessing in finite space bounded by N and all values needed to generate the triplet can be known in nearly EVERY generated case.

So the question is are these randomly generated cases characteristic of N values so that most can be found in finite search space or is there something that makes these randomly calculated solutions repeat to a common N value so that although the vast majority of calculated N values can be reversed they always map to N=6 or some other member of a small subset of all occurring N values (most of occurring N values are not N=6 for example)?

If the generated N values are randomly spread and do not map to N=6 or some other small set of N values then can other elliptical curves be mapped in the same way and usually solved in finite space in the same way? Are all the elliptical curves rational points cases usually solvable if mapped to a similar set of legendre equations?

How can we identify the solvable cases?

So many questions =)

vector – How can I transform a 3D triangle to be coplanar with another congruent triangle? – How can I superimpose two congruent triangles?

I need to transform an equilateral triangle that is lying flat on the xz plane, to become coplanar with another congruent equilateral triangle. I am developing a simple low-poly space exploration game, and my planet generation system requires this.

Terrain chunks are equilateral triangles rather than squares, and they need to be mapped to the faces of an icosphere.

I have found a stack overflow question that outlines a similar problem, but when I tried to implement their solution in Godot, the results were not what I needed. I may have not correctly implemented it, or it might not be an appropriate solution for my problem.

I lost the code that I had used to implement the solution, but it didn’t work anyway. I honestly have no idea where to start with this problem and would appreciate any help.

vector – Godot – How can I transform a 3D triangle to be coplanar with another congruent triangle? – How can I superimpose two congruent triangles?

I need to transform an equilateral triangle that is lying flat on the xz plane, to become coplanar with another congruent equilateral triangle. I am developing a simple low-poly space exploration game, and my planet generation system requires this.

Terrain chunks are equilateral triangles rather than squares, and they need to be mapped to the faces of an icosphere.

I have found a stack overflow question that outlines a similar problem, but when I tried to implement their solution in Godot, the results were not what I needed. I may have not correctly implemented it, or it might not be an appropriate solution for my problem.

I lost the code that I had used to implement the solution, but it didn’t work anyway. I honestly have no idea where to start with this problem and would appreciate any help.

Does congruent triangles apply to this question?

Two identical rods $$BA$$ and $$CA$$ are hinged at $$A$$. When $$BC=8$$ cm, $$angle BAC=30^circ$$ and when $$BC=4$$ cm $$angle BAC=alpha$$.
Show that
$$cos alpha=frac{6+sqrt 3}8$$

I drew two diagrams and tried finding the length of AC (which is the same as AB). Can I use that to solve the triangle with BC=4 using cos rule?

p is a prime congruent to 1 mod 4, find \$sum_{i=1}^{p-2}(frac{i}{p})\$

p is a prime congruent to 1 mod 4, find $$sum_{i=1}^{p-2}(frac{i}{p})$$ where $$frac{i}{p}$$ is the Legendre symbol. I know that -1 is a quadratic residue mod p is p is a prime congruent to 1 mod 4. Any further hints?

Number theory – Ec prime numbers with exponents that are congruent to 10 ^ m mod 41, but not a multiple of 43

If you concatenate two consecutive Mersenne numbers in the base ten, you also get prime numbers, e.g. B. 1023511.
The exponents lead to a prime number that is not a multiple of 43 and congruent to 10 ^ m $$pmod {41}$$ are $$51$$ and $$51456$$.
Do you think this is just a chance?

51 and 51456 $$equiv pm (2 ^ n-4) pmod {559}$$ for some n?

What is the probability that two random numbers are congruent? $$pm (2 ^ n-4) pmod {559}$$

Cutting matrix into 2 congruent pieces [closed]

I have a matrix with 0 and 1. The 1 is similar to a pattern, a shape.
Cut the shape into 2 congruent pieces and make it in the matrix with 2 and 3.
In C ++ and just keep it pls

nt.number Theory – Why is the Congruent Number Problem open?

So I read about the Congruent Number Problem.

One of the sentences on the subject says how the two things are equivalent: a positive integer $$n$$ be a congruent number and elliptic curve $$y ^ 2 = x ^ 3-n ^ 2 x$$ have a non-trivial rational solution. Later it is said in the notes that the problem of the congruent numbers is still open.

My question now is: does the above result not give a criterion to determine whether a certain positive integer is a congruent number or not? The reason could be that we have no way to know when such elliptic curves have a non-trivial solution and therefore no direct way to know if $$n$$ is a congruent number?
And wait, when we say we're looking for a criterion, what exactly do we mean by that?

P .: I am new to the theory of elliptic curves, so I apologize if the above question about MO's standards seems a bit ignorant or not.

If \$ G \$ is the neighboring Chevalley group, are all \$ G ( mathbb Z) \$ subsets with a finite index congruent?

To let $$G$$ be the adjacent Chevalley group, of which everyone is $$G ( mathbb Z)$$'S finite index subgroups congruence subgroups?
I read a sentence saying: When $$G$$ is the universal Chevalley group and not of type $$SL_ {2}$$ every finite index subgroup of $$G ( mathbb Z)$$ is a congruence subgroup, the author has cited this article in relation to this sentence in French.

I searched for sources and results on this topic that I can learn and quote. Unfortunately, all I could find was in French.