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