Given any two positive integers m and n (n > m), it is possible to form three numbers a, b and c where: $$ a = n^2 - m^2, \ b = 2mn, \ c = n^2 + m^2 $$ These three numbers a, b and c are then known as a "Pythagorean triple" - Junior Cycle Mathematics - Question 7 - 2015

To demonstrate that a triangle with sides of lengths a, b, and c is right-angled, we will use the Pythagorean theorem, which states that for a right-angled triangle: $c^2 = a^2 + b^2$

We find:

• $c^2 = 34^2 = 1156$
• $a^2 = 16^2 = 256$
• $b^2 = 30^2 = 900$

Now, add a^2 and b^2: $a^2 + b^2 = 256 + 900 = 1156$

Since $c^2 = a^2 + b^2$, the triangle is confirmed to be right-angled.

We wish to show that if the sides are given by $n^2 - m^2$, $2mn$, and $n^2 + m^2$, then the triangle is right-angled. We will apply the Pythagorean theorem once again:

1. Start by writing the relevant equations:

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

2. We apply the Pythagorean theorem: $c^2 = a^2 + b^2$

   Hence:
   $(n^2 + m^2)^2 = (n^2 - m^2)^2 + (2mn)^2$ Expanding both sides:

   • Left side:
     $(n^2 + m^2)^2 = n^4 + 2n^2m^2 + m^4$
   • Right side:
     $(n^2 - m^2)^2 + (2mn)^2 = (n^4 - 2n^2m^2 + m^4) + (4m^2n^2) = n^4 + 2n^2m^2 + m^4$

3. We see that both sides are equal:
   $n^4 + 2n^2m^2 + m^4 = n^4 + 2n^2m^2 + m^4$

Thus, by Pythagorean theorem, the triangle is right-angled.