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No, i dont have any mathematical prove that i always get an iteger solution. But my mathematical intution says that i always get an interger solution. If someone have any mathematical prove please post here.

In contest time there was a problem in judges checker program so wrong program also get accepted.

Shahriar_monzoor:
Now I find what stupid mistake I made while writing the special judge (Always reading output from judge output).

Also, where did you get the identity:
Code:
5^3+7^3+9^3+10^3=13^3

Use another program to generate it or a formula?

yes, i use another sample program to generate it.

i think using randomize algorithm is more efficient than generating permutation sequence.

Sorry again for a mistake in previous post. To make the program more efficient choose the value for a[n-1] from i= 1 to 220 where the value of i should chosen for a1 to an-2 as be please read it as To make the program more efficient choose the value for a[n-1] from i= 1 to 220 where the value of i s...

Sorry for writing wrong equation. it is a typing mistake. The equation will be 3an^2+3an+1=a1^3+a2^3+.....+(an-1)^3 To choose the value a i use rand() function. The pseudocode is b=rand() if(b&1) a =2*i+1; else a =2*i+2; Since it is a randomize algorithm ,it seems it takes two much time but i got ac...

for n=2 we dont gets any solution as we know from fermats theorem.
for any other value of n we always gets a solution.

You can solve the problem in two ways. simpliest one is: 1^3+6^3+8^3=9^3 9^3+54^3+72^3=81^3 so, 1^3+6^3+8^3+54^3+72^3=81^3 and so. this for n is odd. when n is even the initial equation is, 5^3+7^3+ 9^3+10^3=13^3 so for n=6 , 5^3+7^3+9^3+10^3=13^3+78^3+104^3=117^3 so for n=100 it will be a big numbe...