₦airaland Forum

Miscellaneous: . . .it just became crystal clear the debate will be a tough one.
Thanks for your understanding.

Humphrey77: .NICE , (:THE SEQUENCE IS A MULTIPLE OF 3 WE HAVE THE SEQUENCE AS 6,39,204,1024,5154,25779. ( RULE MULTIPLY 5 TO THE PREVIOUS TERM AND ADD 9 , TO GET THE NEXT TERM @ ,MADAM JACKPOT YOUR FORMULA DID NOT SATISFY THE RULE THAT GOVERNS THE SEQUENCE) :HAPPY

Humphrey77: MY QUEEN GAVE ME THIS QUESTION YESTERDAY AND AS ME TO COMPLETE IT : I said to my self what am i going to do ? told her that am going to post it to my oga on top : complete these : 6,__,204__,__,25779. (hint they are multiple of 3) : HAPPY

Laplacian: hey, i have more unanswered problems than any user on this thread , few in number theory though (as compared 2 elliptic functions)...here are two interestin ones, i've cracked d first one though, maybe u can generalise my result lik u did 4 alpha maximus, or mayb ur method of solution 'll get 2 b prettier than mine, in any case i'll get 2 learn smthin....subsequently, in all my questions, whenever i refer 2 integers, i mean NON-ZERO integers to avoid the supply of trivial solutions:

1.) find the set of all integer pairs (x, y) such that
x²+y² is divisible by xy+1

2.) if
4ⁿ+2ⁿ+1 is a prime, show that n is some power of 3

Humphrey77: find the missing number s; 5,_,1205,_271205,_ ok madam jackpot give me the solutions na

Humphrey77: ok @ jack pot quickly give me the solution to
the problem within 5mins

doubleDx: Wow, great bruv! Kudos man.... nice work!

@Alpha, I see you.... evil worded problem on rampage

Laplacian: for any confusion, refer 2 page 121

eddyvilla: Mine was when I checked my account balance and saw a msg that say: in honor of Mandela, text 95 to 4100 for Harrysong?s TRIBUTE TO MANDELA. Free for a month. To my greatest surprise MTN debited my account of 50 naira. What is NCC even doing in this country?

Laplacian: ...i 'll assume this statement is not comin 4rm jackpot...i suggest the fool who has jackpot's phone should return it 2 her, because jackpot will not make such a futile comment,,,

Laplacian: for any natural number n, the expression (2-√3)ⁿ can be expressed in the form
d+k√3....now d is the required common difference, the required first term, a=d-2k...

@jackpot...if there are infinite natural numbers, then there are infinite solutions

Laplacian: ;
yes

Humphrey77: GIVEN THE sequence 5,_,1205,_,271205,_. find the missing numbers .

Alpha Maximus: ;DAlright, alright, alright! Thanks for the enlightenment. But I didn't say mine was the ONLY way though. Yours still works but mine is more attractive!

Humphrey77: [ I disagree with the third sequence .( it not an AP) THE sequence 15, 112,127

Alpha Maximus: Is there any correlation between such a triple and its successive/previous triple? Is there a general formula for obtaining such triples? Is there any correlation between the first, second and third terms of consecutive triples of this sort?
For example I discovered an easy way of producing Pythagorean Triples sometime ago, the formula is as follows and is sure-fire : 3n, 4n, 5n....where n is ANY number at all! It always works , try it! Assuming n=10, we'll have a Pythagorean triple of 30,40 and 50. Is this valid? Let's confirm.....30²+40²=50²
900+1600=2500
2500=2500 ....thus , the validity has been authenticated . 'N' can even be a decimal to 13 decimal places and it would still work. Laplacian, do you have any formula of this sort to obtain your cited triples in which the sum of one and the product of any two numbers in such a triple will result in the yield of a perfect square?

Alpha Maximus: Is there any correlation between such a triple and its successive/previous triple? Is there a general formula for obtaining such triples? Is there any correlation between the first, second and third terms of consecutive triples of this sort?
For example I discovered an easy way of producing Pythagorean Triples sometime ago, the formula is as follows and is sure-fire : 3n, 4n, 5n....where n is ANY number at all! It always works , try it! Assuming n=10, we'll have a Pythagorean triple of 30,40 and 50. Is this valid? Let's confirm.....30²+40²=50²
900+1600=2500
2500=2500 ....thus , the validity has been authenticated . 'N' can even be a decimal to 13 decimal places and it would still work. Laplacian, do you have any formula of this sort to obtain your cited triples in which the sum of one and the product of any two numbers in such a triple will result in the yield of a perfect square?

Laplacian: ...integers which satisfy my question in increasing order of the magnitude of their first term include;
1, 8, 15
4, 30, 56
15, 112, 127
56, 418, 780
2911, 21728, 40545
...infinitely more...

яhymΣ_ ȴunα†!¢:
Must you insult?

Laplacian: thanks ma'am...but only non-zero integers are allowed

Laplacian: ...pls i need help with this questoin...either a solution or an idea is welcome...

Three non-zero integers are in arithmetic progression, the product of ANY two when when increased by one is a perfect square...find the numbers......

Laplacian: ...pls i need help with this questoin...either a solution or an idea is welcome...

Three non-zero integers are in arithmetic progression, the product of ANY two when when increased by one is a perfect square...find the numbers......

Calculusf(x):
.i respect you bro,but here is my approach...y=x^x^x...y=x^(x^2).try this with calculator and substitute values like 2 and 3 and check if it's correct...for 2...x^x^x=16 and x^(x^2)=16 try that of 3 also...and y=x^x^x means...y={[(x)^x]^x} and from indices u=(x^a)^b=x^ab...then y=x^(x.x)=x^(x^2)...so from the question y=x^x^x...taken natural log of both sides...lny=ln.x^x^x...lny=xlnx^x...(lnx^x=xlnx)then lny=x.xlnx...lny=x^2lnx...1/y.dy/dx=x+2xlnx...dy/dx=x^x^x{x+2xlnx}