Prove that the sum of consecutive perfect cubes (starting with 1) is always a perfect square.
For example:
1=1
1+8=9
1+8+27=36
Let F(n) = 1^3 + 2^3 + ....+ n^3, say.
Then, from the given example:
F(1) = 1 = (1*2/2)^2
F(2) = 9 = (2*3/2)^2
F(3) = 36 = (3*4/2)^2
This leads us to conjecture that:
F(n) = (n(n+1)/2)^2 ....(#)
Now, for n=t, we have:
F(t) = (t(t+1)/2)^2
-> F(t+1) = (t+1)^3 + F(t)
= (t+1)^3 + (t(t+1)/2)^2
= (1/4)*(t+1)^2*(4t+4+ t^2)
= (1/4)*(t+1)^2*(t+2)^2
= ((t+1)(t+2)/2)^2
Thus, the result holds for n=t+1 iff it holds for n=t.
It has already been shown that the result holds for n=1,2,3.
Therefore, the result (#) holds for any positive integer value of n.
Consequently, the sum of consecutive perfect cubes (starting with 1) is always a perfect square.