341Lets write that down
b2≡-1 mod (4n-1)
4n-1 ≡ -1 mod 4
so?
26 Answers
341Where is the contradiction? If you spell it out, you will see it is the same as what is mentioned in the solution posted previously.
39i say see all posts of prophet sir in mathlinks; they are really useful.
and as sir said u will find the mod 13 use there
http://www.mathlinks.ro/search.php?search_author=hsbhatt
341just take note of the guy torajirou who has replied to my qn there. my suggestion is to take some time and read his posts and blogs. In almost every solution of his, he will attempt to show the context in which the problem arises which is very enlightening.
9yeah theres no harm in not understanding
even i dint understand head or tail of it
itll be clear once u learn higher mathematics .
btw i was amazed at how someone cud guess mod 13 so fast by certain methods
im baffled at the level of maths proficiency
39wat no head or tail dude?
its a well known rocess ( prophet sir had shown this to me)
11@ Cele,
There's some reason for taking modulo 13....(thoufh I cud get no head or tail of the reason....posted in the following link)
http://www.mathlinks.ro/viewtopic.php?t=220349
Prophet sir used this once....
341This is by no means infinite descent.
There is simply no way you can demonstrate that this process continues. In fact the very next step will require l to be of different parity than the other quantity.
For infinite descent you will have to show that if (a,b) is a solution set that minimises some +ve integral f(a,b) you can produce (a',b') that further reduces this function.
11Thanx for pointing out the flaw sir....I shall try to rectify it...Edited.
341NT is real fun and some of the results they get are so unbelievably cute. If you have the time try and read some of the good books - Burton's to start off with, then there's Ivan Niven's (this has challenging probs and is real quality stuff). But, if you want to really enjoy this subject read up Sierpsinki's collection of problems. There are some vignettes there, which should leave you smiling.
Then there is the awesome project called PEN http://www.mathlinks.ro/index.php?f=456
This one really hooked me. Read up the solutions to the first few probs (in A3 the technique of Vieta Jumping comes in). See the section on Irrational numbers. It makes you yearn to know more
Sorry if I went overboard, but this is truly fascinating country :D
9ho went thru ur solution sir
very nice
but its way above my knowledge
when did u learn all this sir ?
i thought for a while bhargav posted a Q i cud solve , but it luks like another tuffie inaccessible to 12th pass guys
i wonder if someone cud post an elementary solution ?
@ #9 of b555
what the heck modulo 13 , who wud think like that !!!!
9yup b555s Qs are generally infinity times tuffer than ur Qs sir :P
but he does rarely post some solvable Qs
infact majority of unanswered Qs in this forum is b555's
Chalo b555 keep postin tuffies :P
341There's a small mistake that needs to be corrected.
The second factor is x^2+2x+4
It is of the form 4k+3 as x^2+2x+4 = (x+1)^2+3 and x+1 is even
39anyways that was a great work form u sir. actually taking it with the legrange interpolation is something i could not think off.
truly superb.
as the title says; HATS OFF TO YOU
39another solution: NOT MINE
for any inters x,y we have:
x^3=={0,1,5,8,12} (mod 13)
y^4=={0,1,3,9} (mod 13)
so x^3+y^4== 7 (mod 13)
341BTW, if you are wondering how I seemingly pulled out this one out of my hat, it is modelled after http://www.mathlinks.ro/viewtopic.php?t=67251
The trick is to split it into powers appropriately. then its plain-sailing.
341I remember that question by soumik. I had to appeal to mathlinks for that one. It was inconsistent modulo 13 for Chrissake!!!!!!!!!!
341You can write the equation as x^3 - 8 = -(1+y^4)
Now, if x is even, y has to be odd.
But, then LHS is divisible by 8, whereas RHS is of the form 8k+2, a contradiction.
Now (x-2)(x^2+4x+4) = -(1+y^4)
Now, according to Lagrange, the equation x^2 + 1 \equiv 0 \pmod{p} has no solution if p is of the form 4k+3.
This is quite obvious as
x^2 \equiv -1 \pmod{p} \Rightarrow (x^2)^{\frac{p-1}{2}} \equiv (-1)^{\frac{p-1}{2}} \equiv -1 \pmod{p} \\ \\ \Rightarrow x^{p-1} \equiv -1 \pmod{p}
which is something Fermat would hate to see happen. In Number Theory language -1 is not a quadratic residue mod a prime of the form 4k+3.
Further, if a number is of the form 4k+3, it necessarily has a prime divisor of the form 4k+3 (easily proved)
But if x is odd then x^2+4x+2 \equiv 3 \pmod{4} which means y4+1 has a prime divisor of the form 4k+3 which is a contradiction again.
Hence the given equation has no solution in integers.
(Phew! After long time a prob by b555 that I could solve!)
11Once I had posted a similar qsn....(in another site).....qsn was to prove x^3+y^4=19^19 has no solutions.....it turned out to be tough for even the experts...believe me!
9go on soumink
@b555
(i saw this, luks like we need to consider odd,even cases and proceed )
11Just an intuition.... Take some k such that k^2=y^4... So if I apply congruencies....The only combination is when k^2 leave a remainder 7 and x=3a (modulo 9)... So y=(9n-5)... Meaning eqn gets transformed to the form 9a+9b=-618....which obviously has no solutions... (Well this is not a solid base to support the proof......experts pls hold on before giving the soln for a few more days...)