What two positive intergers cannot be part of a Pythagorean triple?

Answer 1

All pythagorean triples are of the form:
m(a²-b²), 2mab, m(a²+b²)
for positive integers a,b,m where a,b are distinct.

The minimum value of the first term is 1*(2²-1²) = 3
The minimum value of the second term is 2*1*2*1 = 4
The minimum value of the third term is 1*(2²+1²) = 5

Therefore, 1 and 2 cannot be part of a pythagorean triple.

For any other even number, we can use the 2mab term.

For any other odd number, first represent it as m*p where p is prime. Then p=a²-b² = (a-b)(a+b). Let a be one larger than b so that a-b=1 and 2a-1=p or a=(p+1)/2.

Answer 2

That would be 1 and 2, as Quadrillerator showed.
Note that 4 is a member of the triple 3,4,5.
All other positive integers are part of Pythagorean
triples and, in fact, can be used as the smallest member
of a Pythagorean triple.
Why?
Well, if n = 2m +1 is odd(n >=3)
then 2m+1, 2m²+2m, 2m² + 2m +1 form
a Pythagorean triple.
For example, we get the triples
3,4,5
5,12,13
7,24,25
etc.
Next, if n is even but contains an odd factor m,
let q = n/m, find the triple containing
n as above and multiply each member by q
to get a triple with n as smallest member.
Example: 6 contains 3 as a factor,
6/3 = 2 so multiply each of 3,4,5 by 2
to get the new triple 6,8,10.
Finally, suppose n = 2^k(k >= 3)
Then 2^k, 2^(2k-2)-1, 2^(2k-2)+1
form a Pythagorean triple.
Examples:
8,15,17
16,63,65.
32,255,257,
etc.

Answer 3

1 and 2 ,the rest can fit in somewhere in a Pythagorean triple
1, 2, and 3 dont match up

Answer 4

1 and 2 seeing 1^2 +2^2 = 5 therefore hyptoeneuse would be sqrt5

So 1 and 2