Puzzle relating squares to 2020 and 2021 and the question when this happens again? Take the numbers of the current and next year $(2020,2021)$
I noticed that  $ 2020 = 2* 1010 $ and if we take the square of theses divisors added by $2021$ we get the squares $45$ and $1011$.
Or also $$  2021 = 45^2-2^2 $$
and  $$ 2021 = 1011^2 - 1010^2 $$
What is the next year, if there is one, where it will happen that for both squares of the divisor of the old year plus the new year are square?
 A: Let $c$ be any odd integer $3$ or greater, let $b=(c^2-5)/2$, and consider the numbers $2b$ and $2b+1$. We have $2^2+(2b+1)=c^2$, and $b^2+(2b+1)=(b+1)^2$.
The original example is $c=45$, $b=1010$ giving $2b=2020$, $2b+1=2021$.
Axel's example is $c=47$, $b=1102$ giving $2b=2204$, $2b+1=2205$.
The last previous year pair would have come from $c=43$, $b=922$, $2b=1844$, $2b+1=1845$.
Clearly, one gets an infinity of examples from these formulas.
All of these examples come from using $2$ as one of the divisors of the first year of the pair of consecutive years. It's a harder problem to find examples that don't use $2$ as a divisor.
A: Well I wrote a program, and the answer is $2204 = 2\times1102$.
Most of the times (I have tested for $n \leq 50000$ only) the only divisor is $2$, so you end up with $n = 2 \times (n/2)$. There are some quite remarkable numbers though :$144=12\times 12$ $455=13\times35 $ , $4900 = 70 \times 70$, $5719=43\times 133$, $26676=78\times 342$, $41040=180\times 228$.
See you in $2204$.
