i have no clue where to even start. if anyone can help me, it would be greatly appreciated : whiteknite13@hotmail.com

a = n^2 - m^2
b = 2mn
c = n^2 + m^2

a = n² - m²
b = 2·n·m
c = n² + m²

Since a > 0, it must be n > m. Since b is even, even a at the same time would lead to a Pythagorean triple that is a multiple of some lower Pythagorean triple and we will skip those. Consequently, n and m cannot be both even or both odd.

n = 2, m = 1
a = 2² - 1² = 4 - 1 = 3
b = 2·2·1 = 4
c = 2² + 1² = 4 + 1 = 5

n = 3, m = 2
a = 3² - 2² = 9 - 4 = 5
b = 2·3·2 = 12
c = 3² + 2² = 9 + 4 = 13

n = 4, m = 1
a = 4² - 1² = 16 - 1 = 15
b = 2·4·1 = 8
c = 4² + 1² = 16 + 1 = 17

n = 4, m = 3
a = 4² - 3² = 16 - 9 = 7
b = 2·4·3 = 24
c = 4² + 3² = 16 + 9 = 25

n = 5, m = 2
a = 5² - 2² = 25 - 4 = 21
b = 2·5·2 = 20
c = 5² + 2² = 25 + 4 = 29

n = 5, m = 4
a = 5² - 4² = 25 - 16 = 9
b = 2·5·4 = 40
c = 5² + 4² = 25 + 16 = 41
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Find expressions for m, and n in terms of a, b, and c.

c + a = 2n²
c - a = 2m²

n = Ö{(c + a)/2}
m = Ö{(c - a)/2}

Alternately:

c + b = (n + m)²
c - b = (n - m)²

Ö(c + b) = n + m
Ö(c - b) = n - m

n = 1/2·{Ö(c + b) + Ö(c - b)}
m = 1/2·{Ö(c + b) - Ö(c - b)}
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If you are given 3 numbers and asked to find the corresponding values of m and n, how can you decide which number is a, b, and c?

Since c > a and c > b, c is the largest integer of the triple.
Since (c + a)/2, (c - a)/2, c + b , and c - b are all complete squares:

(c + a)/2 = n²
(c - a)/2 = m²
c + b = (n + m)²
c - b = (n - m)²

where n and m integers, neither of c + a, c - a, (c + b)/2, or (c - b)/2 can be a complete square:

c + a = 2n²
c - a = 2m²
(c + b)/2 = (n + m)²/2
(c - b)/2 = (n - m)²/2

Try to add both the remaining numbers to c and/or to subtract both the remaining numbers from c. b is the one that makes a complete square when added to or subtracted from c.
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Find values of m and n for the Pythagorean triple 56, 90, 106.

Since 106 > 56 and 106 > 90, c = 106

106 + 56 = 162 is not a complete square
106 - 56 = 50 is not a complete square
106 + 90 = 196 = 14² is a complete square
106 - 90 = 16 = 4² is a complete square

a = 56
b = 90

n = Ö{(c + a)/2} = Ö{(106 + 56)/2} = Ö(162/2) = Ö81 = 9
m = Ö{(c - a)/2} = Ö{(106 - 56)/2} = Ö(50/2) = Ö25 = 5

Alternately

n = 1/2·{Ö(c + b) + Ö(c - b)} = 1/2·{Ö(106 + 90) + Ö(106 - 90)} =
= 1/2·(Ö196 + Ö16) = 1/2·(14 + 4) = 18/2 = 9

m = 1/2·{Ö(c + b) - Ö(c - b)} = 1/2·{Ö(106 + 90) - Ö(106 - 90)} =
= 1/2·(Ö196 - Ö16) = 1/2·(14 - 4) = 10/2 = 5

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Find values of m and n for the Pythagorean triple 48, 55, 73.

Since 73 > 48 and 73 > 55, c = 73

73 + 48 = 121 = 11² is a complete square
73 - 48 = 25 = 5² is a complete square
73 + 55 = 128 is not a complete square
73 - 55 = 18 is not a complete square

a = 55
b = 48

n = Ö{(c + a)/2} = Ö{(73 + 55)/2} = Ö(128/2) = Ö64 = 8
m = Ö{(c - a)/2} = Ö{(73 - 55)/2} = Ö(18/2) = Ö9 = 3

Alternately

n = 1/2·{Ö(c + b) + Ö(c - b)} = 1/2·{Ö(73 + 48) + Ö(73 - 48)} =
= 1/2·(Ö121 + Ö25) = 1/2·(11 + 5) = 16/2 = 8

m = 1/2·{Ö(c + b) - Ö(c - b)} = 1/2·{Ö(73 + 48) - Ö(73 - 48)} =
= 1/2·(Ö121 - Ö25) = 1/2·(11 - 5) = 6/2 = 3