# Is squaring easier than multiplication? [duplicate]

Let $$T_1(n)$$ be the time complexity of computing the square of an $$n$$-bit integer, and let $$T_2(n)$$ be the time complexity of computing the product of two $$n$$-bit integers.

Assuming that addition is asymptotically faster than multiplication, which of the following is correct?

1. $$T_1(n) = \Theta(T_2(n))$$.
2. $$T_1(n) = o(T_2(n))$$.
3. $$T_2(n) = o(T_1(n))$$.

Please choose one correct option of above .

• Please be more formal in (or elaborate on) addition is asymptotically faster than multiplication. Apr 1, 2021 at 17:35
• So you will live with answers implying ignoring constant factors used in CS more than not. Which is entirely in line with big-O and Little-o relations. Apr 3, 2021 at 7:36
• Please do not edit your question to remove its content. Thank you!
– D.W.
Apr 3, 2021 at 21:18

Observe that $$ab=\frac{1}{2}\left((a+b)^2-a^2-b^2\right)$$,
• This shows me that you do not really understand the definitions. What does it mean for $T_1$ to be the complexity of squaring? Can you tell why $T_1,T_2$ are both $\Omega(n)$? Can you tell that $T_1=O(T_2)$? Apr 1, 2021 at 10:21