Johnny vs Pythagoras?

Question

When teacher explained Pythagorean theorem to students, Johnny complained that the matter is unnecessarily overcomplicated.

'It should be simply c² = a + b' Johnny told the teacher.
'Take for example a right triangle with sides 3,4,5.
My equation correctly yealds 4 + 5 = 9 = 3² '.

'It's not always, true, Jonny' replied the teacher.
'Take for example 5,12,13: 12 + 13 = 25 = ... Uhmm...'

How many more integer side examples can Johhny
produce in support of his theory?

Answer 1

Johnny got his letters wrong; he should have said that his theorem is
b + c = a^2 and not a + b = c^2..

Assuming that revision, there are an infinite number; in fact, every one in which the smallest of the three sides is odd and the other two sides differ by 1.
7, 24, 25
9, 40, 41
11, 60, 61
13, 84, 85,
and so on.

Answer 2

Pythagorean triples are expressed as
k(a²-b²), 2kab, k(a²+b²)

You are asking when does:
k(a²+b²) + k(a²-b²) = 4(kab)²
or
k(a²+b²) + 2kab = k²(a²-b²)²

Let's look at the first one:
2ka² = 4ka²b²
1 = 2b²
Looks grim. No solutions here


k(a²+b²) + 2kab = k²(a²-b²)²
(a+b)² = k(a²-b²)²
So k had better be a square, m²
giving
a+b = m(a²-b²)
or
1 = m(a-b)
From this m=1 so k=1
and a-b = 1 so a=b+1.

Thus, the following pythagorean triples satisfy Johnny:
2b+1, 2b(b+1), 2b²+2b+1


Actually, Isaacs way of looking at it can be refined:
Suppose that a,b and b+1 form a Pythagorean triple. Then we must have:
a² + b² = (b+1)²
a² = 2b + 1, which makes them a Johnny triple

Furthermore, from my development above, we see that the largest two terms of all Johnny triples differ by 1. Therefore, the Johnny triples are exactly those Pythagorean triples where the largest two sides differ by 1 (and it follows that the smaller side will be odd).

Answer 3

a² = b + c ------(1)
c² = a² + b² --------(2)

c² - b² = b + c => (c-b)(c+b)=b+c

either b+c = 0 or c -b = 1 , b,c are positive integers so b+c is not equal to 0

c-b = 1 or c = b+1
so the pythagorean triplets become [sqrt(2b+1),b,b+1]

For a positive integer k :
a = sqrt(2k+1)
b = k
c = k+ 1

This example is possible when 2k+1 is a prefect square for an integer k which is true is k is for the form 2z^2 -1/2 for all integers z

Then the triplets become :
a = 2z
b = 2z^2 -1/2
c = 2z^2 +1/2
for all integers z > 0

So answer is INFINITELY MANY

Answer 4

None. Even the first example is an invalid argument because it does not specify that the c^2 side must be the hypotenuse. His simplification of the theorem omits a crucial premise and is therefore not an equivalent hypothesis.

Answer 5

a lot.

any integers falling into this pattern will work:

n, (n^2-1)/2, (n^2+1)/2

pythagoras:

n^2 + ((n^2-1)/2) ^2
n^2 + (n^4 - 2n^2 + 1)/4 =
n^4 / 4 + n^2/2 + 1/4

((n^2+1)/2) ^2 =
n^4/4 + n^2/2 + 1/4

Jonny:

(n^2-1)/2 + (n^2+1)/2 = n^2

Answer 6

Infinite, this is a shortcut for finding odd number pythagorean triples