Trina's Triangles

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.

1. Investigate Trina's conjecture for several pairs of integers. Does her trick appear to work in all cases, or only in some cases?
2. 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.
3. 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."

Solution

1. To investigate the conjecture, we can construct a table in which we choose two integers $m$ and $n$, and look at the sum of the squares, the difference of the squares, and twice the product of $m$ and $n$: $$ \begin{array}{|c|c|c|c|c|} \hline m & n & m^2 + n^2 & m^2 - n^2 & 2mn \\ \hline 2 & 1 & 5 & 3 & 4 \\ 2 & 2 & 8 & 0 & 8 \\ 3 & 1 & 10 & 8 & 6 \\ 3 & 2 & 13 & 5 & 12 \\ 4 & 1 & 17 & 15 & 8 \\ \hline \end{array} $$ In most cases, the trick seems to work; the triples (3, 4, 5), (6, 8, 10), and others given in the table are familiar Pythagorean triples. However, when $m = n = 2$, we end up with a triple containing the number zero. Since we cannot have a triangle with a side of length zero, the trick does not always work. However, we suspect that it might work as long as we make sure the three numbers generated are all positive.
2. In order to ensure that the three numbers generated are positive, we restrict to the case in which $m$ and $n$ are positive integers, and $m > n$. Also, we change the language of the conjecture slightly to reflect that the numbers are lengths of the sides of a right triangle, not the sides themselves:

   Suppose that $m$ and $n$ are positive integers such that $m > n$. Then the numbers $m^2 + n^2$, $m^2 - n^2$, and $2mn$ are the lengths of the sides of a right triangle.

   To prove this, it suffices to prove that when these three numbers are squared, one square is the sum of the other two. In our table, it appeared that $m^2 + n^2$ was always the largest of the three numbers, so we conjecture that $$ (m^2 + n^2)^2 = (m^2 - n^2)^2 + (2mn)^2. $$ Expanding the left side, we obtain $$ (m^2 + n^2)^2 = m^4 + 2m^2 n^2 + n^4. $$ Expanding the right side, we obtain \begin{eqnarray*} (m^2 - n^2)^2 + (2mn)^2 & = & m^4 - 2m^2 n^2 + n^4 + 4m^2 n^2 \\ & = & m^4 + 2m^2 n^2 + n^4. \\ \end{eqnarray*} Because these two expressions are identical, we have proven that $$ (m^2 + n^2)^2 = (m^2 - n^2)^2 + (2mn)^2. $$ Therefore, of the three numbers generated, the square of one is always the sum of the squares of the other two. Furthermore, we know that all three numbers are positive. In particular, $2mn$ is positive because both $m$ and $n$ are positive; and $m^2 - n^2$ is positive because $m > n > 0$. Therefore, by the converse of the Pythagorean Theorem, the three numbers are the lengths of the sides of a right triangle.
3. To find a triangle satisfying the given requirements, we try using Trina's trick on the integers $m = 13$ and $n = 6$. We get $m^2 + n^2 = 205$, $m^2 - n^2 = 133$, and $2mn = 156$. By the reasoning given above, we know that there is a right triangle with sides of length 133, 156, and 205. Furthermore, the numbers 133, 156, and 205 have no common factors.