New {function, criterion, theorem, conjecture} for discriminant of congruence

[Uncwilly]

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[Zcyyu]

Originally Posted by Uncwilly

Formal note from a moderator:
Do not create any more threads that cover the same subject.
Stop posting the same content repeatedly.
This is your only warning.

I'm so sorry! I don't know how to supplement or modify the original article.

[Uncwilly]

You have a limited time from when you make a post to modify it. If you need to make corrections after that you might quote just the part that needs fixing and then indicate that the you are posting a change. And then state the update.

[Zcyyu]

Originally Posted by xilman

If a man says something and there is no woman around to hear it, is he still wrong?

A new discovery: We found it altogether 12 congruent numbers，it's order of group is >=32
congruent number order of group The upper bound estimate
80451910 Have certain that it is a group of 32 orsders 32
99015854 Have certain that it is a group of 32 orsders 32
124245334 Have certain that it is a group of 32 orsders 32
144362005 Have certain that it is a group of 32 orsders, 128,The order could be > 32
251741886 Have certain that it is a group of 32 orsders 32
253375590 Have certain that it is a group of 32 orsders 32
277872990 Have certain that it is a group of 32 orsders 32
433300054 Have certain that it is a group of 32 orsders 32
565004406 Have certain that it is a group of 32 orsders 32
840075621 Have certain that it is a group of 32 orsders 32
904789886 Have certain that it is a group of 32 orsders 32
936966030 Have certain that it is a group of 32 orsders 32

[Zcyyu]

Originally Posted by Zcyyu

yes, It is that "the congruent number problem"
the conjecture that all congruence solutions all to be a group and the order of the group is a power of 2, it may be one of the most important discoveries in the field of millennial congruence.

Here, we publish the solution of the congruent number 157 for the first time
Since 157 is a prime number with modulo 8 and 5, there must be a solution of type (1.1.157.1) scheme
That is, quaternion quadratic equations
Q1) x^2+y^2=157*z^2 (1)
Q4) w^2-x^2=2*y^2 (2)
The general solution of equation (2) is
x = | u ^ 2-2 * v ^ 2 |, y = 2uv, w = u * u + 2 * v * v
Substituting the expression of X, y for (1), 157 * Z ^ 2 = u ^ 4 + 4 * V ^ 4 (3)
When u = 356411, v = 571761, equation (3) holds in the range of positive integers
z=40,75981,25052 x=52,67710,95761
y=40,75981,25052 w=52,67710,95761
last
s-t=x^2=52,67710,95761^2 s+t=w^2=78,08714,68723^2
s=157*z^2=157*5,31566,61805^2 t=y^2 =40,75981,25052^2