## Saturday, December 29, 2012

### More Equations Involving Natural Numbers - Part 2

To my readers, I get such pleasure from writing on this blog, and fully appreciate each and every one of you. I appreciate the comments. Thank you everyone.

Here's to a Happy and Safe New Years Celebration and to a kick ass 2013!

Onto today's post...

Let A, B, X ∈ N. (Where N is the set of natural numbers (counting numbers). The set can include or not include zero, depending in the definition used. For this blog entry, zero is included.)

One great thing about Casio graphing calculators is that you can make mini-programs. On the current generation (fx-9860g II, Prizm, this includes the older models), within the RUN/MAT mode set the calculator in Linear Input mode. Type each command as you would entering commands in Program mode. Separate each command with a colon (:). I will post screen shots from a Casio Prizm to demonstrate the search for solutions.

For each case, I will present base cases (A = 0, B = 0, and A = B = 0).

(I) A² + B² = X³

Let A = 0, then B² = X³. Likewise, if B = 0, then A² = X³. Without loss of generality, let A = 0.

Then B² = X³. Note that
64 = 4³ = 8² ⇒ B=8, X=4
216 = 6³ = 16² ⇒ B=16, X=6
4096 = 16³ = 64² ⇒ B=64, X=16

More solutions are possible.

Now let A ≠ 0 and B ≠ 0. Then solving for A would yield A = √(X³ - B²). Let B range from 1 to int(√X). Note that int is the integer function. The integer function returns the integer part of a number. Example: int(π) = 3.

Below is the instructions on how I found possible solutions on the Casio Prizm. Note I am looking for integer solutions, ignoring all solutions that were not natural numbers.

Some solutions with X = 1 to 20... (A and B can be interchanged for (I))

(X=2, A=2, B=2)
8 = 2³ = 2² + 2² = 4 + 4

(X=5, A=2, B=11)
125 = 5³ = 2² + 11² = 4 + 121

(X=5, A=5, B=10)
125 = 5³ = 5² + 10² = 25 + 100

(X=8, A=16, B=16)
216 = 8³ = 16² + 16² = 256 + 256

(X=10, A=10, B=30)
1000 = 10³ = 10² + 30² = 100 + 300

(X=10, A=18, B=26)
1000 = 10³ = 18² + 36² = 324 + 676

(X=13, A=9, B=46)
2197 = 13³ = 9² + 46² = 81 + 2116

(X=13, A=26, B=39)
2197 = 13³ = 26² + 39² = 676 + 1521

(X=17, A=17, B=68)
4913 = 17³ = 17² + 68² = 289 + 4624

(X=17, A=47, B=52)
4913 = 17³ = 47² + 52² = 2209 + 2704

(X=18, A=54, B=54)
5832 = 18³ = 54² + 54² = 2916 + 2916

(X=20, A=16, B=88)
8000 = 20³ = 16² + 88² = 256 + 7744

(X=20, A=40, B=80)
8000 = 20³ = 40² + 80² = 1600 + 6400

(II) A² + B³ = X³

If A=0, then B=X, because B³ = X³. For B, X ∈ N.

If B=0, then A² = X³. Some solutions are:
A=1, X=1
A=8, X=4

Let A ≠ 0 and B ≠ 0. Solving for A yields A = √(X³ - B³). Here is the algorithm I used with the my Casio Prizm:

Looking for integer solutions for X=1 to 20, yielded none. Apparently, solutions only exist when either A or B is allowed to be zero.

(III) A³ + B³ = X²

If A=0, then B ³ = X ². Likewise, if B=0, then A³ = X².

Without loss of generality, let A = 0. Some solutions are:

(A=0, B=4, X=8)
64 = 4³ = 8²

(A=0, B=6, X=16)
216 = 6³ = 16²

(A=0, B=16, X=64)
4096 = 16³ = 64²

Now assume, once again, that A ≠ 0 and B ≠ 0. Using the same strategies as (I) and (II), let's solve for A. Then A = ∛(X² - B³). Here is the algorithm I used with the Casio Prizm:

The only solution that I found with X = 1 to 20 is:

(A=1, B=2, X=3). [A and B can be interchanged]
9 = 3² = 1³ + 2³ = 1 + 8

Stay safe and take care, Eddie. Thanks everyone.

This blog is property of Edward Shore. 2012