Help with an AoPS Geometry problem

[skimomma]

I think I am missing something.  I have the solution manual but I cannot understand part of the rationalization for the approach.

 

This is exercise 6.3.6 dealing with Pythagorean triples.

 

After setting up the problem we are to solve for a and b in the equation:

 

a^2 + b^2 = (97)^2

 

This is rearranged to:

 

a^2 = (97+b)(97-b)

 

where we are to use trail and error to find b so that the right hand side of the equation results in a perfect square.

 

The solution manual says that (97+b) and (97-b) must each (on their own) be perfect squares and that is key to solving the problem.  Why is this true and where does that "rule" come from?

 

I have reworked the problem and numbers using a simple 3-4-5 triple where I do not find this to be true:

 

a^2 = (5+b)(5-b) 

 

If a=4 and b=3, the equation is satisfied and does result in a triple but both (5+b) and (5-b) are NOT perfect squares so I cannot figure out how we were to come to the conclusion that this must be true in order to solve for 6.3.6.

 

 

[RootAnn]

I have reworked the problem and numbers using a simple 3-4-5 triple where I do not find this to be true:

 

a^2 = (5+b)(5-b) 

 

If a=4 and b=3, the equation is satisfied and does result in a triple but both (5+b) and (5-b) are NOT perfect squares so I cannot figure out how we were to come to the conclusion that this must be true in order to solve for 6.3.6.

 

I can't speak to this exactly, but if b=4, then (5+b) and (5-b) are both perfect squares.

 

Which makes me wonder if a has to be assigned to the shortest side? 

 

5-12-13

a=5, b=12, c=13

a^2 = (13+b) (13-b) --> works

 

if a=12, b=5, c=13

a^2 = (13+b) (13-b)  --> doesn't work

[Twigs]

…

 

I have reworked the problem and numbers using a simple 3-4-5 triple where I do not find this to be true:

 

a^2 = (5+b)(5-b) 

 

If a=4 and b=3, the equation is satisfied and does result in a triple but both (5+b) and (5-b) are NOT perfect squares so I cannot figure out how we were to come to the conclusion that this must be true in order to solve for 6.3.6.

 

but if you let a=3 and b=4, it does work.

 

a^2 = (5+b)(5-b) 

3^2 = (5+4)(5-4) 

9=(9)(1)

9=9

 

(65, 72, 97)

65^2 = (97+72)(97-72)

4225=(169)(25)

4225=4225

 

It looks like a has to be the smallest number and c has to be the largest number.

[Ravi B]

This is rearranged to:

 

a^2 = (97+b)(97-b)

 

where we are to use trail and error to find b so that the right hand side of the equation results in a perfect square.

 

The solution manual says that (97+b) and (97-b) must each (on their own) be perfect squares and that is key to solving the problem.  Why is this true and where does that "rule" come from?

 

  I don't think the solutions manual says that 97 + b and 97 - b must each be perfect squares.  My copy of the manual says: "Since (97 - b)(97 + b) must be a perfect square, we try to find values of b that make both 97 - b and 97 + b perfect squares."  In other words, it's merely saying that if 97 - b and 97 + b are perfect squares, then their product (97 - b)(97 + b) is a perfect square.  And fortunately, we can find b such that 97 - b and 97 + b are squares.

 

 

Edited November 17, 2017 by Ravi B

[skimomma]

  I don't think the solutions manual says that 97 + b and 97 - b must each be perfect squares.  My copy of the manual says: "Since (97 - b)(97 + b) must be a perfect square, we try to find values of b that make both 97 - b and 97 + b perfect squares."  In other words, it's merely saying that if 97 - b and 97 + b are perfect squares, then their product (97 - b)(97 + b) is a perfect square.  And fortunately, we can find b such that 97 - b and 97 + b are squares.

 

 

 

Ah.  So they don't both have to be perfect squares on their own but by looking for perfect squares we can be assured their product will be a perfect square.  That makes sense.  Thanks!  However, they is no way my 14yo was going to figure that one out on her own.  Sigh.

[Ravi B]

Yes, it's a challenging problem, which is why it's starred and has a hint at the back of the textbook.  If you or your 14 yo is interested in further properties of Pythagorean triples, check out Euclid's formula on the Wikipedia page:

https://en.wikipedia.org/wiki/Pythagorean_triple

For example, we can use Euclid's formula to solve the problem you posed here (find a Pythagorean triple with hypotenuse 97).  Using Euclid's formula, we just need a way to express 97 (not 97^2) as a sum of two squares, which is easy (9^2 + 4^2).  Once we do that, we can determine the two legs: 9^2 - 4^2 = 65 and 2*9*4 = 72.