Prove that if P is the perimeter of a Pythagorean Triangle with integral sides a, b and c, then P divides ab. (Hint: Use the formula on the bottom of page 74 for the description of Pythagorean triples)

Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
Chapter7: Locus And Concurrence
Section7.1: Locus Of Points
Problem 47E: In Exercises 39 and 42, refer to the line segments shown. Use the following theorem to construct a...
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I added the entire work to get that formula from that book, the page is attached. Thank you. 

Prove that if P is the perimeter of a Pythagorean Triangle with integral sides a,
b and c, then P divides ab. (Hint: Use the formula on the bottom of page 74
for the description of Pythagorean triples)
Transcribed Image Text:Prove that if P is the perimeter of a Pythagorean Triangle with integral sides a, b and c, then P divides ab. (Hint: Use the formula on the bottom of page 74 for the description of Pythagorean triples)
last equation can be expressed as
(a + B)(a - B) = 1.
wwww
All the numbers concerned are rational, so if the product of two numbers is 1, they are
O reciprocals. That is, one number must be m/n and the other n/m, where m and n are
t integers. Setting
a + B =
and
a - B =
we nd by addition that
m
and by subtraction that
1
1 (m
B =
Consequently,
m² + n²
m2 -n²
wwww
(1)
2mn
2mn
But y = Bx and z = ax; if we now put x = 2mn, so as to get a solution in integers, it
follows that
=2mn.
y = m-n².
z = m? + n2.
=%²
r
wwwww
These are well-known formulas for nding right triangles with sides of integral length and
were used in Hellenistic times by Diophantus (circa 150), the most original mathematician
of late antiquity.
Transcribed Image Text:last equation can be expressed as (a + B)(a - B) = 1. wwww All the numbers concerned are rational, so if the product of two numbers is 1, they are O reciprocals. That is, one number must be m/n and the other n/m, where m and n are t integers. Setting a + B = and a - B = we nd by addition that m and by subtraction that 1 1 (m B = Consequently, m² + n² m2 -n² wwww (1) 2mn 2mn But y = Bx and z = ax; if we now put x = 2mn, so as to get a solution in integers, it follows that =2mn. y = m-n². z = m? + n2. =%² r wwwww These are well-known formulas for nding right triangles with sides of integral length and were used in Hellenistic times by Diophantus (circa 150), the most original mathematician of late antiquity.
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