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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 ...

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 ...

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 ...