More Equations Involving Natural Numbers - Part 2

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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!

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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).

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(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

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(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.

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(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

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Stay safe and take care, Eddie. Thanks everyone.


This blog is property of Edward Shore. 2012

More Equations Involving Natural Numbers

Greetings!

For the following let a, b, and n ∈ N. That is all variables are natural numbers (the counting numbers). Note that the set of natural numbers may or may not include 0. I will include 0 for this blog entry. This blog entry is all about trying to find solutions to equations - in an analytic manner.

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a + b = a * b

If a = b, then:
a + a = a * a
2a = a^2

The only solution to this is a=2. (2*2 = 2^2 = 4)

Allowing for possibilities that a and b are different:
a + b = a * b
a + b - a * b = 0
a * (1 - b) = -b
a = -b/(1 - b) = b/(b - 1) = 1 + 1/(b - 1)

The only way that 1/(b - 1) ∈ N is when b = 2. Then a = 2.

The only solution to a + b = a * b is a = b = 2.

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a^2 + b^2 = a * b

Let's say if a = b. Then:

a^2 + a^2 = a * a
2a^2 = a^2
a^2 = 0
a = 0

This implies that b=0. This works because 0^2 + 0^2 = 0 * 0 = 0.

What if we allow the possibility that a ≠ b?

Let's start with subtracting a*b from both sides:
a^2 + b^2 - a*b = 0
a^2 * (1 - b/a) + b^2 = 0

This lead me to believe that b is a multiple of a. Let b = a * n. Then:

a^2 * (1 - n) + a^2 * n^2 = 0

Assuming a ≠ 0,
(1 - n) + n^2 = 0

Which leads to
n = (1 ± i √3)/2.

Not good for searching for solutions such that a,b ∈ N.

In the general case:
a^2 + b^2 - a*b = 0
(1)(a^2) - (b)(a) + (b^2) = 0

Then:
a = (b ± √(b^2 - 4b^2))/2 = (b ± i b √3)/2 = b * (1 ± i √3)/2 which is not a natural number.

The only solution to a^2 + b^2 = a*b is a=b=0

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a + b^2 = a * b

Let's start with subtracting a*b from both sides:
a + b^2 - a*b = 0
(1)(b^2) - (a)(b) + (a) = 0

Which leads to:
b = (a ± √(a^2 - 4a))/2 = a/2 ± 1/2 * √(a^2 - 4a)

This implies that:
(1) a^2 - 4a must be a perfect square,
(2) a^2 - 4a must be even, and
(3) a must be even.

If a=2: √(2^2 - 4*2) = √(-4) = 2i. So a=2 is not a solution.

If a=4: √(4^2 - 4*4) = 0 which leads to b = 4/2 ± 0 = 2.
Then with a=4 and b=2:
4 + 2^2 = 4 * 2
8 = 8

A solution is found! Are there any more?

If a = 6, then a^2 - 4a = 12, not a perfect square.
If a = 8, then a^2 - 4a = 32, not a perfect square.
If a = 10, then a^2 - 4a = 60, not a perfect square.
If a = 12, then a^2 - 4a = 96, not a perfect square.
If a = 14, then a^2 - 4a = 140, not a perfect square.
If a = 16, then a^2 - 4a = 192, not a perfect square.

If a^2 - 4a is a prefect square, then a^2 - 4a = n^2. Then:
a^2 - 4a - n^2 = 0.
Then a = 2 ± √(4 + 2n^2). This is inconclusive.

One solution to a + b^2 = a * b is a=4 and b=2. There could be more.

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a^2 + b^2 = 2 * a * b

Let's solve the equation in terms of b:
a^2 - 2 * a * b + b^2 = 0
(1)(b^2) - (2 * a)(b) + (a^2) = 0

And:
b = (2 * a ± √(4 * a^2 - 4 * a^2))/2 = (2 * a ± √0)/2 = a

This makes sense because when a=b, a^2 + a^2 = 2 * a * a = 2 * a^2.

Testing a few solutions:

a=3 and b=3:
3^2 + 3^2 = 18 and 2*3*3 = 18

a=6 and b=6:
6^2 + 6^2 = 72 and 2*6*6 = 72

The solution to a^2 + b^2 = 2*a*b is when a = b

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Enjoy!

Eddie



This blog is property of Edward Shore. 2012

Equations without Whole Number Solutions - Fruitless Search?

Hi everyone! Hopefully you are fine today. Not long until 2013. And yes, I am 99.44% confident that the human race will still be on Earth be here come 12/22/2012.

x + x^2 = n^2

Let n,x ∈ N. N represents the natural numbers. Natural numbers are commonly referred to the whole numbers 1, 2, 3, etc. Some mathematicians include 0.

Are there any integer solutions to x + x^2 = n^2 with n,x ∈ N?

Our first instinct is most likely to go grab the nearest calculator or computer. Observe that:

(I). x + x^2 = n^2

(II). x^2 * (1/x + 1) = n^2

Taking the (principal) square root of both sides yields:

(III). x * √(1/x + 1) = n

In order for (III) to be true:
1. The quantity 1/x + 1 has to be an integer,
2. 1/x + 1 has to be a perfect square, and
3. x = 1/x + 1

The only natural number that allows condition 1 to be true is when x = 1.

Then 1/x + 1 = 2.

We know that 2 is not a perfect square (√2 ≈ 1.41421), so condition 2 fails. (II) does not fit because 1 * √2 = √2, leaving n = √2, not fitting the requirement that n,x ∈ N.

According to this analysis, there are no natural number solutions to x + x^2 = n^2.

x + x^2 + x^3 = n^q

Let n,x ∈ N. Suppose q = 2. Can we find solutions with these conditions?

x + x^2 + x^3 = n^2

x^2 * (1/x + 1 + x) = n^2

x * √(1/x + 1 + x) = n

Once again the only way 1/x + 1 + x is an integer is that when x = 1. However when x=1, 1/x + 1 + x = 3, and √3 ≈ 1.73205. And 1 * √3 = √3, n = √3, which is not a natural number.

There are no solutions (in the natural number set) for x + x^2 + x^3 = n^2.

What about q = 3?

Then x + x^2 + x^3 = n^3

x^3 * (1 + 1/x + 1/(x^2)) = n^3

x * ∛(1 + 1/x + 1/(x^2)) = n

Again, the only possible candidate is when x=1, but that leaves n = ∛3. No solutions in the natural number set.

Not all is "lost"...

Consider trying to find solutions to:

x + x^2 + x^3 + x^4 = n^2 where n,x ∈ N.

Then x * √(1/x + 1 + x + x^2) = n.

If x = 1, then √(1/x + 1 + x + x^2) = √4 = 2, and n = 1 * 2 = 2. Success! I believe x=1 and n=2 is the only solution to this equation with these conditions imposed.


Happy Holidays everyone and see you next time!

Eddie

This blog is property of Edward Shore. 2012