Solve the following Diophantine equation

Question

I need help with solving the following diophantine equation: $$x^2+y^2=2018$$

Thanks a lot in advance!

@Dr.SonnhardGraubner I think you miscalculated something $49^2=2401$ — Vinyl_cape_jawa, Feb 18 '19 at 20:09
To speed the (already simple) search, note that you can just solve the problem for $1009$ instead of $2018$ and then use the fact that $1^2+1^2=2$ to "build" a solution for $2018$. — lulu, Feb 18 '19 at 20:09
$x$ and $y$ must both be odd and less than $45$. Just try the possible values for $x$ and see which ones work. — saulspatz, Feb 18 '19 at 20:10
By the 'sum of two squares theorem' $$$$ An integer $n$ greater than $1$ can be written as a sum of two squares $a^2+b^2$ if and only if its prime decomposition contains no prime $p\equiv 3\mod 4$ raised to an odd power.$$$$
We know that $2018=1009^1·2^1$. — Dr. Mathva, Feb 18 '19 at 20:12

score 0 · Answer 1 · answered Feb 18 '19 at 20:10

0

HINTS:

$$2(31^2)<2018<2(32^2)$$ $$44^2<2018<45^2$$

Use these two to narrow down the range of possibilities, and test them by assigning all $y$ and seeing if $2018-y^2$ is a square number.

answered Feb 18 '19 at 20:10

Rhys Hughes

1 Answers1