hats off to one who solves this

I wish the JEE had NT in its syllabus....

Lets write that down

b²â‰¡-1 mod (4n-1)

4n-1 ≡ -1 mod 4

so?

Sorry, sorry bhat sir, I've messed it up all....

Where is the contradiction? If you spell it out, you will see it is the same as what is mentioned in the solution posted previously.

i 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

just 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.

yeah 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

wat no head or tail dude?

its a well known rocess ( prophet sir had shown this to me)

@ 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....

This 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.

Thanx for pointing out the flaw sir....I shall try to rectify it...Edited.

thank u sir for the inputs :)

no takers?

cele , didnt see this one?

NT 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

ho 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 !!!!

yup 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

There'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

anyways 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

BTW, 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.

I remember that question by soumik. I had to appeal to mathlinks for that one. It was inconsistent modulo 13 for Chrissake!!!!!!!!!!

You 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!)

Once 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!