## Task

Alice was having a conversation with her friend Trina, who had a discovery to share:

“Pick any two integers. Look at the sum of their squares, the difference of their squares, and twice the product of the two integers you chose. Those three numbers are the sides of a right triangle.”

Trina had tried this several times and found that it worked for every pair of integers she tried. However, she admitted that she wasn't sure whether this "trick" always works, or if there might be cases in which the trick doesn't work.

- Investigate Trina's conjecture for several pairs of integers. Does her trick appear to work in all cases, or only in some cases?
- If Trina's conjecture is true, then give a precise statement of the conjecture, using variables to represent the two chosen integers, and prove it. If the conjecture is not true, modify it so that it is a true statement, and prove the new statement.
- Use Trina's trick to find an example of a right triangle in which all of the sides have integer length, all three sides are longer than 100 units, and the three side lengths do not have any common factors.

## IM Commentary

This task is a fleshing-out of the example suggested in A-APR.4 of the Common Core document, using the polynomial identity $(x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2$ to generate Pythagorean triples.

Students must investigate Trina's conjecture to discover that it does not work in all cases; in particular, the trick fails if the two chosen integers are the same, or if one of the integers is zero, since in these cases one of the sides of the triangle given will have length zero. This means that students must attend to precision when writing a corrected version of the conjecture, being careful to restrict to (for example) the case in which the two integers chosen are positive and distinct.

Proving the corrected conjecture requires some polynomial arithmetic and provides an opportunity to check whether students square binomials correctly, avoiding the common error of stating that $(x + y)^2 = x^2 + y^2$. It also provides an opportunity to review the logical distinction between the Pythagorean Theorem and its converse (8.G.6); the converse is what is needed to prove this conjecture.

The condition that the side lengths have no factors in common in part (c) is designed to prevent students from coming up with the desired right triangle by "scaling up" a smaller triangle. For example, students couild take a standard 3-4-5 right triangle and scale by a factor of 100 to find a 300-400-500 right triangle, which is clearly not in the spirit of the problem. Students may or may not discover that for the triangles to have side lengths with no common factor, it suffices to ensure that their choice of $m$ and $n$ have no common factor.

It is a remarkable theorem, well beyond the scope of this problem, that in fact every right triangle with integer side lengths can be constructed by Trina's process, i.e., corresponds to some choice of $m$ and $n$. Interested students or teachers should investigate the "classification of Pytghagorean triples."