# All boils down to one

**Number Theory**Level 3

\[\begin{array} \\ 3^2+4^2 &=& 5^2 \\ 5^2+12^2 &=& 13^2 \\ 6^2+8^2 &=& 10^2 \\7^2 + 24^2 &=& 25^2 \\ \end{array} \]

A Pythagoras Triplet \((A,B,C)\) satisfy the condition that \(A,B,C\) are positive integers such that \(A^2 + B^2 = C^2\). From above, we can see that \((3,4,5), (5,12,13), (6,8,10),(7,24,25) \) are all Pythagoras Triplets.

Note that equations above are equivalent to

\[ \begin{array} \\ (\color{blue}2^2 - \color{green}1^2)^2 + (2 \cdot \color{blue}2 \cdot \color{green}1)^2 & = & (\color{blue}2^2 + \color{green}1^2)^2 \\ (\color{blue}3^2 - \color{green}2^2)^2 + (2 \cdot \color{blue}3 \cdot \color{green}2)^2 & = & (\color{blue}3^2 + \color{green}2^2)^2 \\ (\color{blue}3^2 - \color{green}1^2)^2 + (2 \cdot \color{blue}3 \cdot \color{green}1)^2 & = & (\color{blue}3^2 + \color{green}1^2)^2 \\ (\color{blue}4^2 - \color{green}3^2)^2 + (2 \cdot \color{blue}4 \cdot \color{green}3)^2 & = & (\color{blue}4^2 + \color{green}3^2)^2 \\ \end{array} \]

**True or false?**:

"All Pythagoras Triplets \((A,B,C)\) can be written as \( (m^2 - n^2, 2mn, m^2 + n^2 ) \) for positive integers \(m,n\) with \(m>n\)."

Note that the values of \(A\) and \(B\) are interexchangeable, so the triplet \((3,4,5) \) is equivalent to \((4,3,5) \).

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