Prelude to 2021

Observe the following structure of polygonal numbers $P(s,n)$ :

Source: http://i1.wp.com/mathandmultimedia.com/wp-content/uploads/2012/08/polygonal-numbers.png

The $n$ th $s$ -gonal number can be represented as a "polygon" with side length $n$ , a single dot at one corner (red), $s-1$ "sticks" of length $n-1$ (blue), and $s-2$ "triangles" of dimension $n-2$ (yellow), where each triangle represents a triangular number.

Each triangular number of dimension $n-2$ is the sum of the first $n-2$ positive integers. Hence:

$1+2+3+\dots+n-2=\frac{1}{2}(n-2)(n-1)$

Therefore,

$P(s,n)=1+N_{sticks}+N_{triangles}=1+(s-1)(n-1)+(s-2)\left[\frac{1}{2}(n-2)(n-1)\right]=\frac{(s-2)n^2-(s-4)n}{2}$

Since $\min\limits_{b\text{ prime}} (a+b)(a-b)=2021$ and $a=P(s,n) \in \mathbb{Z}^+$ , then:

$(a+b)(a-b) \geq 2021$

$a^2 \geq b^2 + 2021$

$a \geq \sqrt{b^2 + 2021} > b > 0$

Therefore, $a+b > a-b \in \mathbb{Z}^+$ . We then convenient notice that $2021 = 43 \times 47 = 2021 \times 1$ :

$a+b = 47$ $a-b = 43$ $(a,b) = (45,2)$

$a+b = 2021$ $a-b = 1$ $(a,b) = (1011,1000)$

...in which the latter does not satisfy the condition of $b$ being prime. Note as well that since a distinct solution exists that satisfies the given conditions, there can no longer be any other solution for $(a,b)$ .

We then look at the possible values of $s$ and $n$ and yield the following solutions: $P(3,9)$ , $P(6,5)$ , $P(16,3)$ , $P(45,2)$ .

Therefore,

$S = 3+6+16+45 = 70$ $\Phi(S) = 70 \left(1-\frac{1}{2}\right) \left(1-\frac{1}{5}\right) \left(1-\frac{1}{7}\right) = 24$ $S+\Phi(S)+\sqrt{\Phi(S)} = 98.899\dots$

$\sum_{i,j \leq S+\Phi(S)+\sqrt{\Phi(S)}} i+j = \sum_{i,j \leq 98.899\dots} i+j = (2+3)+(3+5)+\dots+(83+89)+(89+97)\\ = \boxed{2021}$