Let us take the numbers 2, 3 and 4, 3 being the multiple of 3...
4^2 - 2^2
= 16 - 4
= 12
We can see that the result of the calculation (12) is equal to the original number (3) multiplied by 4.
To further explain this, let us write the proof in terms of algebra.
X-1, X, X+1
(X+1)(X+1) = X^2 + 2X + 1
(X-1)(X-1) = X^2 - 2X + 1
(X^2 + 2X + 1) - (X^2 - 2X + 1) = 4X
Given a number "X" in the 3 times table, the difference of the squares of the numbers "X-1" and "X+1" (the numbers consecutively before and after X) is equal to 4X.
Trying this with the 5 times table, it turns out that the equation stays the same, irrelevant of the number.
Amended Conclusion : Given a number "X", the difference of the squares of the numbers "X-1" and "X+1" (the numbers consecutively before and after X) is equal to 4X
EXTENSION:
X-2, X, X+2
(X-2)(X-2) = X^2 - 4X + 4
(X+2)(X+2) = X^2 + 4X + 4
(X^2 + 4X + 4) - (X^2 - 4X +4) = Answer
Answer = 8X
Final Proof:
Let the 3 numbers be A, B and C.
B-D, B, B+D
(B-D)(B-D) = B^2 - 2D + D^2
(B+D)(B+D) = B^2 + 2D + D^2
(B^2 + 2D + D^2) - (B^2 - 2D + D^2)
= 4D
Hence proved.