Question

- resulted from prims subject :

- a right triangle can be formed by length sides of prims ?

what mean that need being true the pythagora's
c^2 = a^2 +b^2
for a,b and c prim numbers
- any idea,opinion ?

Answer 1

@ganeshie8

Answer 2

@Kainui

Answer 3

so than not - any proof of this why ?

Answer 4

all 3 numbers would have to be odd. Is this possible?

Answer 5

Oh - I suppose one of them could be 2.....

Answer 6

why all 3 - one from these can being 2 - so what is even - or not ?

Answer 7

yea i agree.
Its a tricky one , this.

Answer 8

I would think its highly unlikely one is 2 . If you see a list of these triples none of the early ones have a 2 in them and when we get to the higher numbers a 2 is not possible

Answer 9

if a and b are both odd we have
(2n + 1^2 + (2n + 1)^2 = 8n^2 + 8n + 2 which must be even

Answer 10

is that valid?

Answer 11

- oh - only if both numbers are the same , of course! lol

Answer 12

however (2n + 1)^2 + (3n + 1)^2 would also be even...

Answer 13

on this website i have got the right explination and the proof of this my question
- https://www.mathsisfun.com/numbers/pythagorean-triples.html

Answer 14

\[c^2 = a^2 +b^2 \implies a^2 = (c+b)(c-b)\]

Since \(a\) is prime, we get two cases :
Case 1) \(c+b = c-b = a\)
Case 2) \(c-b=1\) and \(c+b=a^2\)

Both happen to be impossible.

Answer 15

@ganeshie8 yes this is nice very and easy understanding - i agree it

Answer 16

so and thank you

Answer 17

@phi your opinion please - ? thank you in advance

Answer 18

Awesome! This reminds me of a particularly fun problem you might enjoy @jhonyy9

What are all the primes p, q, and r that satisfy this equation:
\[p+q^2=r^2\]

Answer 19

sorry but i think you missed the exponent of p - or not ?