Problem Solving Question

[jayden0002]

Hi, I recently joined so I'm not sure if I'm posting this to the right place but here is my question.
If Two positive integers, differing by 2, are such that the difference of their squares is a perfect square. Find several pairs of these integers and give a general rule for finding such pairs. Can you please help and explain!!! Thanks
Edit: This is a Problem solving question our teacher gave with no actual textbook. I am currently in Year 10 (Australia) doing Year 10 Advanced Maths. Thanks for the answers to the problem, will get to work now! Thank you all!

 

[JeffM]

First, thank you very much for giving the problem exactly and completely.

Second, please read a summary of our guidelines here

https://www.freemathhelp.com/forum/threads/112086-Guidelines-Summary?p=433156&viewfull=1#post433156

As you will have seen in the guidelines , we want to know what you are studying and what work you have already done.

In this case, the question itself identifies a method. It asks FIRST for a number of numerical examples. You can find the examples by systematic numerical experimentation. (This is easiest with a computer if you have any programming skills.)

\(\displaystyle 1^2 + (1 + 2)^2 = 1 + 9 = 10.\) 10 is not a perfect square.

\(\displaystyle 2^2 + 4^2 = 4 + 16 = 20.\) 20 is not a perfect square.

\(\displaystyle 9 + 25 = 34.\) No.

\(\displaystyle 16 + 36 = 52.\) No.

\(\displaystyle 25 + 49 = 74.\) No.

\(\displaystyle 36 + 64 = 100.\) Yes.

The problem requires you to find at least one more example, but, before you do, you might want to ponder for a few moments what might explain your first example other than pure accident.

 

[Dr.Peterson]

Hi, I recently joined so I'm not sure if I'm posting this to the right place but here is my question.
If Two positive integers, differing by 2, are such that the difference of their squares is a perfect square. Find several pairs of these integers and give a general rule for finding such pairs. Can you please help and explain!!! Thanks

JeffM showed a way to find numbers so that the sum (rather than difference) of the squares is a square. You should try doing the same.

Then you might want to try using algebra. Suppose the smaller of the two numbers is x. What would the larger number be? What expression would represent the difference of their squares? Then you can think about how you might use this to find values of x for which the difference is s perfect square (say, n^2).

Let us know what you find.

 

[HallsofIvy]

Hi, I recently joined so I'm not sure if I'm posting this to the right place but here is my question.
If Two positive integers, differing by 2, are such that the difference of their squares is a perfect square. Find several pairs of these integers and give a general rule for finding such pairs. Can you please help and explain!!! Thanks

Call the two integers x and y with x the larger. Since they differ by 2, x- y= 2. The difference of the squares is a perfect square so x^2- y^2= 4, or x^2- y^2= 9, or x^2- y^2= 16, etc. Notice that x^2- y^2= (x- y)(x+ y) so, since x- y= 2, those equations can written 2(x+ y)= 4, 2(x+ y)= 9, 2(x+ y)= 16, etc. One thing that tells us is that the square must be even so we must have x+ y= 4/2= 2, x+ y= 16/2= 8, x+ y= 36/2= 18, etc. But again, they differ by 2: y= x- 2 so x+ y= x+ x- 2= 2x- 2. That is, we must have x+ y= 2x- 2=2 so 2x= 4 and x= 2, y= 2- 2= 2 (which is not a positive integer so this is NOT a solution) or x+ y= 2x- 2= 8 so 2x= 10, x= 5 and y= 5- 2= 3, or x+ y= 2x- 2= 18 so 2x= 20, x= 10 and y= 10- 2= 8, etc.