A set of square numbers

Lets look at $M$ befor the process: $M = \left\{1; 2; 3; 4; \cdots; n \right\}$

Thus ${ S }_{ M} =\sum _{ i=1 }^{ n }{ i } =\frac { n\cdot \left( n+1 \right) }{ 2 }$

Now look at $N$ befor the process: $N = \left\{ { 1 }^{ 2 }; { 2 }^{ 2 }; { 3 }^{ 2 }; { 4 }^{ 2 }; \cdots ; { n }^{ 2 } \right\}$

Thus ${ S }_{ N }=\sum _{ i=1 }^{ n }{ { i }^{ 2 } } =\frac { n\cdot \left( n+1 \right) \cdot \left( 2\cdot n+1 \right) }{ 6 }$

Now you have to delete all numbers from $M$ which appear in $N$ as well. But which numbers appear in $N$ as well?

Imagine this example for $n=10$ :

$M = \left\{1; 2; 3; 4; 5; 6; 7; 8; 9; 10 \right\}$

$N=\left\{ { 1 }^{ 2 }; { 2 }^{ 2 }; { 3 }^{ 2 }; { 4 }^{ 2 }; { 5 }^{ 2 }; { 6 }^{ 2 }; { 7 }^{ 2 }; { 8 }^{ 2 }; { 9 }^{ 2 }; { 10 }^{ 2 } \right\} =\left\{ 1; 4; 9; 16; 25; 36; 49; 64; 81; 100 \right\}$

As you can see, all numbers after ${3}^{2}$ are not in $M$ . So we delete all square numbers from $M$ which are smaller or equal $n$ . We know we have a formula for the first $n$ square numbers (see ${ S }_{ N }$ ). But how many squares are less or equal to our $n$ ?

${ 3 }^{ 2 }\le 10\le { 4 }^{ 2 }$

${ 3 }\le \sqrt { 10 } \le { 4 }$

Thus the number of squares less than or equal to $n$ is the biggest integer, less than or equal to $\sqrt { n }$ . This leads us to the definition of the floor function . Thus the sum of all squares less than or equal to $n$ is:

${ S }_{ squares\le n } = \sum _{ i=1 }^{ \left\lfloor \sqrt { n } \right\rfloor }{ { i }^{ 2 } } =\frac { \left\lfloor \sqrt { n } \right\rfloor \cdot \left( \left\lfloor \sqrt { n } \right\rfloor +1 \right) \cdot \left( 2\cdot \left\lfloor \sqrt { n } \right\rfloor +1 \right) }{ 6 }$

Cancel out all squares, less than or equal to $n$ , from $M$ is the same as subtract ${ S }_{ squares\le n }$ from ${ S }_{ M}$ .

${S}_{afterProcess} = { S }_{ M} - { S }_{ squares\le n } =\sum _{ i=1 }^{ n }{ i } - \sum _{ i=1 }^{ \left\lfloor \sqrt { n } \right\rfloor }{ { i }^{ 2 } } = \frac { n\cdot \left( n+1 \right) }{ 2 } - \frac { \left\lfloor \sqrt { n } \right\rfloor \cdot \left( \left\lfloor \sqrt { n } \right\rfloor +1 \right) \cdot \left( 2\cdot \left\lfloor \sqrt { n } \right\rfloor +1 \right) }{ 6 }$

This is the generalization of $S$ .

Thus :

${ S }_{ { 2019 }^{ 2 } }=\frac { { 2019 }^{ 2 }\cdot \left( { 2019 }^{ 2 }+1 \right) }{ 2 } -\frac { \left\lfloor \sqrt { { 2019 }^{ 2 } } \right\rfloor \cdot \left( \left\lfloor \sqrt { { 2019 }^{ 2 } } \right\rfloor +1 \right) \cdot \left( 2\cdot \left\lfloor \sqrt { { 2019 }^{ 2 } } \right\rfloor +1 \right) }{ 6 }$

${ S }_{ { 2019 }^{ 2 } }=\frac { { 2019 }^{ 2 }\cdot \left( { 2019 }^{ 2 }+1 \right) }{ 2 } -\frac { 2019\cdot \left( 2019+1 \right) \cdot \left( 2\cdot 2019+1 \right) }{ 6 }$

${ S }_{ { 2019 }^{ 2 } }=\frac { { 2019 }^{ 2 }\cdot \left( { 2019 }^{ 2 }+1 \right) }{ 2 } -\frac { 2019\cdot 2020\cdot 4039 }{ 6 }$

${ S }_{ { 2019 }^{ 2 } }=8305616109871$

Put simply, the problem wants us to find the sum of all numbers from $1$ to $2019^2$ and then subtract the numbers $1^2, 2^2, 3^2, ..., 2019^2$

There are two very famous formulas for these specific operations:

$1 + 2 + ... + n = \dfrac{n(n+1)}{2}$

$1^2 + 2^2 + ... + n^2 = \dfrac{n(n + 1)(2n + 1)}{6}$

So the sum we want is

$\dfrac{2019^2(2019^2+1)}{2} - \dfrac{2019(2019 + 1)(2(2019) + 1)}{6} = 8305616109871$

In an sixteenth of a second, $\text{Total}\left[\text{Complement}\left[\text{Table}\left[n,\left\{n,2019^2\right\}\right],\text{Table}\left[n^2,\{n,2019\}\right]\right]\right] \Rightarrow 8305616109871$ , direct computation.