# Pythagorean Triangles

2. If p>3, show that p divides the sum of its quadratic residues that are also least residues.

(see attached file for diagram)

4. Here is a quadrilateral, not a parallelogram, with integer sides and integer area:

(a) What is its area?

(b) Such quadrilaterals are not common; can you find another?

(c) Could you find 1,000,000 more?

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1. (a) Prove that if p≡7("mod" 8), then p | (2^(((p-1))⁄2)-1).

(b) Find a factor of 2^83-1.

Solution:

(a) Suppose p≡7 ("mod" 8). We want to show that 2^(((p-1))⁄2)≡1 ("mod" p).

We can now make use of the following theorem.

Theorem 1. If p is an odd prime, then

(2⁄p)=1 "if " p≡1 "or" 7("mod" 8),

(2⁄p)=-1 "if " p≡3 "or" 5("mod" 8).

Since p≡7 ("mod" 8), it follows by the Theorem 1 that (2⁄p)=1. Therefore, by Euler's Criterion, we have 2^(((p-1))⁄2)≡1 ("mod" p).

(b) Let p=2∙83+1=167. Since 167 is prime and p≡7 ("mod" 8), it follows from part (a) that 167 is a factor of 2^83-1.

2. If p>3, show that p divides the sum of its quadratic residues that are also least residues.

Solution: Let's first introduce the following theorems.

Theorem 2. If p is an odd prime, then there are exactly ((p-1))⁄2 nonzero quadratic residues (mod p) and ((p-1))⁄2 quadratic non-residues (mod p).

Proof: See the notes at the end.

Theorem 3. Every prime p has ϕ(p-1) primitive roots.

Proof: See Dudley's Elementary Number Theory, Section 10.

Now suppose that p>3 is a prime. By Theorem 3, p has a primitive root. Let g be a primitive of p. Then g is a least residue modulo p and the order of g ...

#### Solution Summary

This solution helps with questions regarding quadratic reciprocity and pythagorean triangles.