A Pythagorean triple is an ordered triple of positive integers(a,b,c) such that a^2+B^2=c^2.Find
the number of the Pythagorean triples such that all the numbers a,b and c are prime?
for any natural numbers m and n where m and n are relatively prime
2mn, m^2-n^2 and m^2 + n^2 will form a Pythagorean triple.
since the smallest triple is 3,4,5,
one of the numbers (the 2mn) will always be even.
So there are no such Pythagorean triples where all 3 are primes.
To find the number of Pythagorean triples where all the numbers are prime, we need to look for prime numbers that satisfy the Pythagorean theorem: a^2 + b^2 = c^2.
First, we need to consider the nature of Pythagorean triples. In a Pythagorean triple, two of the numbers are always even (if one number is odd, the other two numbers will also be odd, making their squares odd). So, one of the prime numbers in the triple must be 2.
Now, let's consider the value of c. Since c is the hypotenuse of the right-angled triangle, it must be greater than both a and b. However, if c is a prime number, it cannot be even since the only even prime number is 2. Therefore, c must be an odd prime number.
Since a prime number greater than 2 is always odd, a and b must be both odd or one odd and the other equal to 2 (a = 2 and b = odd, or a = odd and b = 2).
For the case where a = 2 and b = odd, we know that c must be an odd prime number greater than b. There are infinitely many odd prime numbers, so for each odd prime number greater than 2, we have a Pythagorean triple (2, odd, c).
For the case where a = odd and b = 2, we also know that c must be an odd prime number greater than a. Again, there are infinitely many odd prime numbers, so for each odd prime number greater than 2, we have a Pythagorean triple (odd, 2, c).
Therefore, the number of Pythagorean triples where all the numbers are prime is infinite.
In conclusion, there is an infinite number of Pythagorean triples where all the numbers (a, b, c) are prime.