A Parallelogram Has Sides of Lengths 39 and 25 and a Diagonal of Length 34. So, What Makes It So Special!

Thanks to TC's inspired challenge to our readers in a comment on the Medians of a Triangle post, I've decided to expand it into an investigation for our readers and students (geometry with some trig needed).

Consider a parallelogram whose sides have lengths 39 and 25 and with one diagonal of length 34.

(a) Explain why this parallelogram is unique, i.e., all parallelograms with these characteristics are congruent. Why was it not necessary to specify that the 'shorter' diagonal was given?

(b) A parallelogram has sides of lengths a and b and diagonals of lengths c and d. Use the Law of Cosines to show that
c² + d² = 2(a² + b²).

(c) Determine the length of the other diagonal. As an alternative, how would you do it without the formula in (b)?

(d) Determine the area of this parallelogram.

(e) So what makes this parallelogram unusual?

Comments:
(1) From the comments on the Medians post (Cotton Blossom and others), we know that we can construct rectangles and rhombuses whose sides and diagonals have integer lengths, but the above demonstrates a parallelogram that is neither of these special cases.
(2) The formula in (b) is not too difficult to prove, however, finding solutions to this Diophantine equation or a general solution is far more challenging!
(3) Note that the parallelogram in this challenge also has integral area. Finding other such parallelograms is not a simple exercise!

M^2 - N^2 = 12...Prove there is only one solution in positive integers and much more

Update:
(1) See the visualization for the difference of squares posted on 1-9-08.
(2) Read the comments in this post for considerable clarification and instructor guidelines and suggestions. Mathmom's and Eric's comments are particularly insightful.

This post can be developed into an activity for prealgebra through first-year algebra students (or even 2nd year algebra). The last part is more challenging.
The focus here is on developing a method/strategy that can be used to solve similar Diophantine equations. The other objective is to introduce the ideas and methods of proof. This problem may later be used to solve a recent math contest problem for which I obtained permission to discuss on this blog. I am fully aware that many students will 'solve' these equations by Guess-Test methods, but they need to go further.

STUDENT/READER PROBLEMS/ACTIVITIES

(a) Prove there is only one solution in positive integers for the equation:

M² - N² = 12

Note: If we omit the word positive, what would the solution(s) be?

(b) Determine all positive integer solutions:

M² - N² = 15

(c) Determine all positive integer solutions:

M² - N² = 36

(d) Let's investigate for what positive integer values of P, M² - N² = P has NO solutions in positive integers.

(i) Determine at least 5 positive integer values of P for which the above equation has no positive integer solutions.

(ii) (More challenging) Describe all values of P for which the above equation has no solutions. Justify your result.

Note: All students should have success with (i), although some may struggle to find 5 values. Part(ii) should challenge the student who has finished the other parts in rapid order and sits there complacently!

Additional Comment: If P is itself a perfect square, our equation is obviously related to the most famous equation in geometry. Thus, if P = 9 or P = 16, for example, students should recognize something! For this reason you may want to have students consider these values when doing this investigation. More to come...