MHB Proof of Quadratic Reciprocity: Undergrad Friendly

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The discussion focuses on finding an elementary proof of the Law of Quadratic Reciprocity suitable for first or second-year undergraduates studying number theory. One participant suggests a proof involving counting lattice points within a square, referencing a theorem related to the Legendre symbol. They emphasize that while the proof may seem standard, it is crucial for understanding the concept of quadratic reciprocity. The conversation highlights the importance of visualizing lattice points to derive the reciprocity law effectively. Overall, the thread serves as a resource for students seeking accessible mathematical proofs.
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Does anyone know of an elementary proof of the Law of Quadratic Reciprocity. I am looking for a proof that 1st or 2nd year undergraduate can understand. Searching for a proof which is digestible for a student who is doing a first course in number theory.
 
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Sure. The proof I know of is by counting lattice points of a square. I thought it was pretty standard, but giving it here in case anyone who reads does not know :

Theorem : $$\left(\frac{a}{p}\right) = (-1)^{\sum_{j = 1}^{(p-1)/2} \left \lfloor \frac{a \cdot j}{p} \right \rfloor}$$

There is a pretty standard proof of this and can be found in any introductory NT text, so I am omitting this here; but do not think that this result is trivial! It's actually the key to the proof of QR.

Now, consider this

View attachment 2160

There are exactly $\left \lfloor \frac{q \cdot j_1}{p} \right \rfloor$ and $\left \lfloor \frac{p\cdot j_2}{q} \right \rfloor$ integer points on the boldfaced straightlines, respectively. And there are $(p-1)/2$ and $(q-1)/2$ integer points on the x and y-axis respectively, so let $j_1$ and $j_2$ run through those. Hence the number of lattice points are

$$\sum_{j_1 = 1}^{(p-1)/2} \left \lfloor \frac{q \cdot j_1}{p} \right \rfloor + \sum_{j_2 = 1}^{(q-1)/2} \left \lfloor \frac{p \cdot j_2}{q} \right \rfloor$$

But then simply counting the lattices gives $(p-1)/2\cdot (q-1)/2$ lattice points in the square. Hence exponentiating both sides to $-1$ gives

$$\left(\frac{p}{q}\right) \left(\frac{q}{p}\right) = (-1)^{(p-1)/2\cdot (q-1)/2}$$
 

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