I am trying to find all integers $m$ such that $m$ is relatively prime to 30, and $m=6x^2+5y^2$ for some integers $x,y$. Note that we must have: $y$ is odd, $(y,3)=1=(x,5)$. Using these conditions, I was able to show that $m=11$ or $29$ mod $30$. So I was first trying to handle the case $m=11$ mod 30. Note that $11=6\cdot1^2+5\cdot1^2$, but $41$ is not of the form $6x^2+5y^2$. So I was wondering, for which $n$ do we have $30n+11=6x^2+5y^2$? Also, for such $n$, can we obtain $x$ and $y$ satisfying $30n+11=6x^2+5y^2$?
2026-03-30 09:51:14.1774864274
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For what integers $n$, do we have $30n+11=6x^2+5y^2$ for some integers $x,y$?
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if n>0, then left hand side is positive, and 30n+11 chooses an ellipse from right-hand side which contains an ellipse for each value in $\{ (x,y) | f(n,x,y)=g(n,x,y), (n,x,y) \in (Z,Z,Z) \}$ where $f(n,x,y)=30n+1$ and $g(n,x,y)=6x²+5y²$. After you realize that it's just an ellipse for positive values of the left hand side, then the next step is to realize that the integers are a grid that needs to intersect the ellipse.
The values to look for are mod 120, specifically $11, 29, 59, 101 \pmod{120}.$ There is a fairly clean theory about primes that are represented by a binary form. Since the class number $h(\mathbb Z[\sqrt{-30}]) = 4$ but there is one form in each genus, the primes $p = 5 x^2 + 6 y^2 $ are described precisely by congruences.
Other than the primes $2,3,5,$ the primes such that $(-120|p) = 1$ are represented as such:
$$ \begin{array}{ccccccc} 1. & 1,&31,&49,&79, & \pmod{120}& :: \; \; x^2 + 30 y^2 \\ 2. & 17,&23,&47,&113, & \pmod{120}& :: \; \; 2x^2 + 15 y^2 \\ 3. & 13,&37,&43,&67, & \pmod{120}& :: \; \; 3x^2 + 10 y^2 \\ 6. & 11,&29,&59,&101, & \pmod{120}& :: \; \; 5x^2 + 6 y^2 \\ \end{array} $$
Products of such primes are represented by one of the four forms, specified by Gauss composition. For example, $11 \cdot 29 = 319 = 17^2 + 30 \cdot 1^2 = 7^2 + 30 \cdot 3^2.$ Suppose we call the forms 1,2,3,6, where the last one is for $6 y^2 + 5 x^2.$ For a product of such "good" primes, write each one with a code 1,2,3,6. Then multiply, table $1^2 = 2^2 = 3^2 = 6^2 = 1.$ Next $2 \cdot 3 = 6$ and $2 \cdot 6 = 3.$ Finally $3 \cdot 6 = 2.$ Well, here is a $2 \cdot 3 = 6.$ As in $17 \cdot 13 = 221 = 5 \cdot 5^2 + 6 \cdot 4^2 = 5 \cdot 1^2 + 6 \cdot 6^2 $
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