Congruence

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High School Math Grades 9-12, Ages 13-18

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#1 h Apr 3, 2025, 10:34 AM

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Find the number of integer pairs (x, y) satisfying the congruence equation:

3y² + 3x²y + y³ ≡ 3x² (mod 41)

for 0 ≤ x, y < 41.

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#2 h Apr 4, 2025, 1:39 PM

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Bump bump

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#3 h Apr 6, 2025, 8:42 AM

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Any ideas?

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#4 h Apr 7, 2025, 3:39 AM

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If $y\neq0,1$ , then the original equation asserts that

$\left( 3x(1-y) y^{-1} \right)^2 = 3(1-y)(y+3)$
The number of solutions satisfying this equation is given by

$\left(\frac{3 (1-y)(y+3)}{p}\right) + 1,$
where $\left(\frac{\cdot}{p}\right)$ denotes the Legendre symbol.

Moreover, since

$\sum_{y=0}^{p-1} \left(\frac{(y-1)(y+3)}{p}\right) = -1,$
it follows that the summation for this part is

$\sum_{y=2}^{p-1}\left( \left(\frac{3 (1-y)(y+3)}{p}\right) + 1 \right) = p-2+\left(\frac{-3}{p}\right)\left(\sum_{y=0}^{p-1} \left(\frac{(y-1)(y+3)}{p}\right) - \left(\frac{-3}{p}\right) - \left(\frac{0}{p}\right) \right) = p - 2 - \left(\frac{-3}{p}\right) - 1.$
When $y = 0$ , it is evident that $x = 0$ is the unique solution; and when $y = 1$ , there is evidently no solution. Therefore, the total number of solutions is

$p - 2 - \left(\frac{-3}{p}\right) - 1 + 1 = p - 2 - \left(\frac{-3}{p}\right) = p - 2 - \left(\frac{p}{3}\right).$
Since $p = 41 \equiv 2 \pmod{3}$ , it follows that $\left(\frac{p}{3}\right) = -1$ . Consequently, the number of solutions is

$p - 2 - (-1) = p - 1 = 40.$

This post has been edited 1 time. Last edited by lbh_qys, Apr 7, 2025, 3:41 AM

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