Uotiop69

@Math_files If number is even then divide by 2, if odd then multiply by 3 then +1 add, keep doing recursive with this way to any number, at the end it will reach to 1, That is a fact with no proven maths,

Hit me with the craziest math facts you know.

@trashkunkun @usrbinishan @ParryRamMe @Math_files 0 is not a positive interger, Question said a,b to be a positive interger.

@uotiop69 @usrbinishan @ParryRamMe @Math_files Having 0 as one of the integer will also work. So, again, another result different from your rule. a = 0 and b = any integer a = n and b = n³ Are valid sets that produce infinite solutions. BUT, that's not ALL of them.

This problem is widely remembered as one of the most difficult mathematics problems of its time, originating from the International Mathematical Olympiad. In 1988, when the IMO was held in Australia, the final question—Problem 6—became legendary for its extreme difficulty. Only 11 students managed to achieve a perfect score on it. Even Terence Tao, a 13-year-old prodigy representing Australia at the time and later a Fields Medalist—the highest honor in mathematics—scored just one point on this problem. What makes the story even more remarkable is that when the problem was reviewed by the panel of experts before the exam, even they were unable to solve it. Despite this, it was still included in the competition.

@rb357 @Math_files a=n, b=n^3 is a one family, ( there exists more ), I didn't calculated

@uotiop69 @Math_files But not all the solutions are of the form a = n, b = n^3. E.g. a = 8, b = 30 => ab+1 = 240+1 = 241, a^2+b^2 = 900 + 64 = 964 964/241 = 4, a perfect square

@trashkunkun @usrbinishan @ParryRamMe @Math_files Yes, I said I found one family, not all, there still exists more family

@uotiop69 @usrbinishan @ParryRamMe @Math_files For example, the solution using a = 8, and b = 30 will be valid as ab + 1 = 241 a² + b² = 964 (a² + b²)/(ab + 1) = 4 But, this rule is clearly different from your assumption... That is "b = a³ and the resulting division is a²".

@usrbinishan @ParryRamMe @Math_files That integer is literally a^2 That was the trick 💀

@uotiop69 @ParryRamMe @Math_files the question has the constraint that the division results in an integer: "such that ab + 1 divides a^2 + b^2"

@usrbinishan @ParryRamMe @Math_files Let a = 3 Let b = 4 (3^2+4^2)/(3*4+1) (9+16)/(12+1) (25)/(13) = ???? This is not square, That is counter example of random values can't satisfied something square, Math is about to describe constraints,

@uotiop69 @ParryRamMe @Math_files let's say you asked me to prove that sin^2(t) + cos^2(t) = 1. would you be satisfied with me saying sin^2(t) + cos^2(t) = 1 whenever t is some multiple of pi, since cos^2 will be 1 and sin^2 will be 0?

@thurty33z We back, $clawcode

@uotiop69 Haven't seen you bag work 😕 are we back ?

@usrbinishan @ParryRamMe @Math_files That literally wrong, Question asked to validate answer equal to something square, That something is a^2 ( other type of solution exists as well, not gonna find )

@uotiop69 @ParryRamMe @Math_files you have to show that this works for any a and b, not just some special family with a relation between a and b

@ParryRamMe @Math_files Solution comes from constraints, not all can work, (a^2+b^2)/(ab+1) is not equal to z^2 always, There is a constraints for it to happens ( like a family of the solutions )

@uotiop69 @Math_files (a^2+b^2)/(ab + 1) can be square of any random integer and not necessarily of a

@ParryRamMe @Math_files (a^2+b^2)/(ab + 1) = a^2 a^2 + b^2 = a^3*b + a^2 b^2 = a^3*b b=a^3 It comes from this

@uotiop69 @Math_files Basis for b= n^3 ?

@ParryRamMe @Math_files Let a=n b = n^3 then (n^2 + n^6) / (n^4 + 1) now factor it, (n^2 * ( n^4+1)) / (n^4 +1 ) cancel out (n^4+1) Then n^2 remains, So, for any a=n, and b=n^3, result is proven to be n^2 remains.

@uotiop69 @Math_files That is obvious. Prove it in terms of a and b

@mathsrick_ n=(n-1)*2 + 1 Simple

Reasoning ability test 📚✅

@IQTestBrain a=6 Simple

Find a

@IQTestBrain 4c + 2c + c = 12 6c = 12 c = 2 So, b = 4

@MathGuyTFL x=50 Easy

Find the value of X.

@Math_files Pattern to noticed is, some are a=n, then b=n^3

@Math_files a=2, b=8 a=3, b=27 a=4, b=64 a=5, b=125 a=6, b=216 a=7, b=343 a=8, b=2 a=8, b=30 a=8, b=512 a=9, b=729 a=27, b=3 a=27, b=240 a=30, b=8 a=30, b=112 a=64, b=4 a=112, b=30 a=112, b=418 a=125, b=5 a=216, b=6 a=240, b=27 Many solutions, Lol,

@Mr_Relaxy Was doing research on some math and computation theories.

@agnesrium @realsigridjin $gajae means nothing! Betting on ClawCode, not random animal coins which have zero value. ( Might pump if cabals push, don't care )

@uotiop69 @realsigridjin lol just ape $gajae it is his token