We want to prove that in a Pythagorean triple $(a, b, c)$, where $a^2 + b^2 = c^2$, if one of $a$, $b$, or $c$ is odd, then two of the numbers must be odd and one must be even.

Step 1: Definitions of Odd and Even Using Modulo

• A number $n$ is even if $n \equiv 0 \ (\text{mod} \ 2)$, meaning $n$ is divisible by 2.
• A number $n$ is odd if $n \equiv 1 \ (\text{mod} \ 2)$, meaning $n$ leaves a remainder of 1 when divided by 2.

Step 2: Analyze the Pythagorean Triple Modulo 2

For a Pythagorean triple, we have: $a^2 + b^2 = c^2$ We want to analyze this equation modulo 2.

Case 1: All numbers are even

If all of $a$, $b$, and $c$ are even, then:

• $a \equiv 0 \ (\text{mod} \ 2)$
• $b \equiv 0 \ (\text{mod} \ 2)$
• $c \equiv 0 \ (\text{mod} \ 2)$

Substituting this into the equation $a^2 + b^2 = c^2$: $0^2 + 0^2 = 0^2 \quad \text{(mod 2)}$ This holds true, but this case doesn't satisfy our condition that one of the terms is odd.

Case 2: Two numbers are even and one is odd

Now, assume two numbers are even, and one is odd. Without loss of generality, assume $a$ and $b$ are even, and $c$ is odd:

• $a \equiv 0 \ (\text{mod} \ 2)$
• $b \equiv 0 \ (\text{mod} \ 2)$
• $c \equiv 1 \ (\text{mod} \ 2)$

Substituting into $a^2 + b^2 = c^2$: $0^2 + 0^2 = 1^2 \quad \text{(mod 2)}$ $0 + 0 = 1 \quad \text{(mod 2)}$ This results in a contradiction, since $0 \neq 1$. Therefore, it is impossible for two numbers to be even and one to be odd.

Case 3: Two numbers are odd and one is even

Now, assume two numbers are odd and one is even. Without loss of generality, assume $a$ and $b$ are odd, and $c$ is even:

• $a \equiv 1 \ (\text{mod} \ 2)$
• $b \equiv 1 \ (\text{mod} \ 2)$
• $c \equiv 0 \ (\text{mod} \ 2)$

Substituting into $a^2 + b^2 = c^2$: $1^2 + 1^2 = 0^2 \quad \text{(mod 2)}$ $1 + 1 = 0 \quad \text{(mod 2)}$ $2 \equiv 0 \quad \text{(mod 2)}$ This is true, so the equation holds. Therefore, two of the numbers can be odd, and one can be even.

Step 3: Conclusion

From the analysis, we have shown that:

• It is impossible for two numbers to be even and one to be odd.
• It is possible for two numbers to be odd and one to be even.

Thus, in a Pythagorean triple, if one of the terms is odd, then exactly two of them must be odd, and one must be even.

Would you like further clarification or a deeper explanation of any part?

Here are some related questions to explore:

1. How does the general formula for Pythagorean triples (in terms of $m$ and $n$) reflect this property?
2. What happens in the case of primitive Pythagorean triples with respect to odd and even terms?
3. Can you find a Pythagorean triple where all terms are odd? Why or why not?
4. What happens to the parity of terms in a non-primitive Pythagorean triple?
5. What would be the result if all terms in the Pythagorean triple were even?

Tip: In a primitive Pythagorean triple, one leg must always be even, and the hypotenuse is always odd.