The year 2020

Number Theory Level 4

Does there exist two positive integers $x$ and $y$ that are not divisible by $505$ and sastify the equation below?

$x^2 + 2019y^2 = 4 \cdot 505^{2020}$

This is part of the series: " It's easy, believe me! "

Cannot be determined. No, there is not.. Yes, there is.

1 solution

Mark Hennings
Sep 17, 2017

Note that $2020 \; = \; 1^2 + 2019\times1^2 \hspace{2cm} 505^3 \; = \; 757^2 + 2019 \times 252^2$ and hence $\begin{aligned} 4 \times 505^{2020} & = \; 2020 \times (505^3)^{673} \; = \; |1 + i\sqrt{2019}|^2 \times \big(|757 + 252i\sqrt{2019}|^2)^{673} \\ & = \; \big|(1 + i\sqrt{2019})(757 + 252i\sqrt{2019})^{673}\big|^2 \; = \; |X + iY\sqrt{2019}|^2 \; = \; X^2 + 2019 Y^2 \end{aligned}$ for some integers $X, Y$ , where $X + iY\sqrt{2019} = (1 + i\sqrt{2019})(757 + 252i\sqrt{2019})^{673}$ .

Moreover, $5$ divides neither $X$ nor $Y$ . Thus it is certainly true that $505$ divides neither $X$ nor $Y$ .

Where did the second equation (505^3=...) come from?

Joe Mansley - 2 years, 2 months ago

We are looking for numbers that can be written as $x^2 + 2019y^2$ . Obviously we have $2020 = 4\times505 = 1^2 + 2019 \times 1^2$ . That leaves us $505^{2019} = 505^{3\times673}$ to handle. On the "it would be great if it worked" principle, I did a computer search to see if $505^3$ could be written as $x^2 + 2019y^2$ . There are only $252$ possible positive values of $y$ , so it was not much of a hunt.

Mark Hennings - 2 years, 2 months ago

The year 2020

1 solution

0 pending reports