what? how is this possible?

i m sure all they re trying to say is: April fool!!!! :):)

lol never realized that totally fooled!!! you got me guys! :P :P

no sample I/O?

No X problem this time around ?

@ Abhijit: No!!! i think they re working on it!!! i read somewhere that this month there are going to be 2 challenge problems!! Check the forums!!!

heyii can you please tell me form of output

whether

output:

3 4

or

output:

3

4

@Santosh: Either is ok!!

Either output will work fine.

@Subhash The problem statement is fine :)

Can you give a sample input/output

I think this is a trick question, we have to give  a boolean answer stating wether such a thing is possible or not.

So it that just April fool?

It's in contradiction with Fermat last theorem..

Or it must have some trick!

@ admins

very funny guys. This is the fermat's last theorom. it states that a^n+b^n=x^n if and only if n<=2

so for all numbers greater than 2, u will never be able to find the answer.

so similarly, x^n-a^n=b^n is not possible for n<=2...

April Fools to all huh?? :P

Provide a Example Input And Output plz..

any test cases, if possible?

Actually I have discovered a truly marvelous code to solve this question but the computational power of your system is too low to run it.

@Neelesh: LOL =))

But there is nothing about a and b constraints. The can be negative, I guess. In that case, solution is always exists. But failed to solve this problem, anyway :(.

@Max

You need to find two positive integers a and b

read question carefully

in input:

take 'x' first or 'n' first

in output:

print 'a' first or 'b' first

pls hlp?

@admin Please provide some sample input output cases

wat is the range of test cases and values of 'a' & 'b'?

@Anurag

they cant provide input/output sample cases

bcoz by doing dat, trick to this ques will get revealed

i think question is perfectly right but don't know where is the mistake must be very small.

you are doing programming not math.

@santosh kumar

but it is the fermat's last theorem:-

a^n + b^n =c^n is not possible for n>2

the above can be re-written as:

c^n - b^n =a^n

which is the eqn given to us for n>=3

1 <= x <= 100000 3 <= n <= 10

is this

1 <= x <= 100000 & 3 <= n <= 10

or 1 <= x <= 1000003 <= n <= 10 which is slightly contradictory :P

I know the solution to this problem!!!

Every test case file begins with '0' (The number of test cases!!).

Output nothing. :)

don't we need a limit on T

@admin: This is terribly unprofessional behaviour.

heyii if we think ^ as bitwise operator then it is possible but didn't get accepted

anyone has any idea about this

sample i/p:

1

6 2

o/p:

10 8

but codechef always ignore my solution by saying "time limit exceeded"

@$um@n

n >= 3, so 2 is invalid.

@$um@n

3 <= n <= 10

I am getting the answer

I/P

1

1000 3

O/P

6459 6451

is it right pls tell me

LOL it's so funny people are actually trying to code this!!

@max lapan

yh,u r right man..i was considering n>=1

@codechef....

Plz tell us if its solution exists or not..I have been thinking on it for 3 hrs.I searched da whole google n they have even given us the proof of Ferment's Last theorem..I definitely think its wrong..Just tell us if it can be solved or not..Its just waste of time giving this problem.Plz reply sooooooon...

i'm almost done with the solution,i just need to know

that

are 'a' and 'b' non zero integers?

am getting the answer

I/P

1

1000 3

O/P

6459 6451

is it right pls tell me

can anyone tell me

@Anurag does not work last digit of 6459^3 is 9 and last digit of 6451^3 is 1 so last digit of 6459^3 - 6451^3 is 8... last digit of 1000^3 is 0 :P

@admin: This 'fools' version is 'symbolic' of the contests organized here.

@admin: This 'fools' version is 'symbolic' of the contests organized here.

@admin: This 'fools' version is 'symbolic' of the contests organized here.

@everyone

If there are multiple solutions, output the one with the smallest a.

least value of a is when

a=x;

this gives b=0; which satisfies the eqn also.

so everytime o/p will be:-

x 0

where x is i/p.

@me

but i got wrong answer through above logic

Lovely Aprill fool prank ! You nearly got me :P

if b can be zero?

@Chirayu

the question says positive numbers

@admin..Learn some algebra before posting problems like this. Read Fermat's last theorem which states that a^n+b^n=c^n is not possible for n>2. Or change the question to find real numbers instead of integers.

@kinshuk "learn algebra" real numbers have too many solutions :P and there is no smallest positive a :P

Just a general comment... Fermat's last Theorem is one conjuncture which was called a "theorem" even before it was proved... and the Poincaré conjuncture is still called a "conjuncture" (it's proved)...

watz wrong wit u guys...?

did anyone hack dis site n driving the users AWAY from this site or wat...?!!!

i had a lot of respect towards this site but now u r loosing that...

hope u will rectify everything...

n y r the previous winners of old contests not working on this contest....?

hey buddies have a look on this point...

@admin: when will the actual contest start?? We are all waiting!!!