Number Theory Question

T

Trenters4325

Junior Member
Joined
Apr 8, 2006
Messages
122
Given that b^2 = c(a+c) and that a,b,c are positive integers whose sum we want to minimze, why does c need to be a square?
 
I don't quite know, but a little tinkering provides some insight.

\(\displaystyle \L\,\frac{b^{2}}{c} = a+c\)

\(\displaystyle \L\,\frac{b^{2}}{c}+b = a+b+c\)

\(\displaystyle \L\,b(\frac{b}{c}+1) = a+b+c\)

This would suggest that we want 'b' as small as possible and 'c' as great as possible.

For what it's worth...
 
I think the idea is that if c is not a square, then you can divide the whole equation by a number greater than one. Why this is true is what I don't understand.
 
Trenters4325 said:
Given that b^2 = c(a+c) and that a,b,c are positive integers whose sum we want to minimze, why does c need to be a square?
Click to expand...
I think you mean "primitive cases";
a=3, b=2, c=1 : 4 = 1(3 + 1) : c is a square
a=6, b=4, c=2 : 16 = 2(6 + 2) : c is not a square
 
I'm not quite sure what the term "primitive case" means, but that seems to be the right idea. We want a, b, and c to be relatively prime.
 
OK, so "primitive", "coprime", and "relatively prime" are all synonymous. But that doesn't really answer my question...
 
Trenters4325 said:
... But that doesn't really answer my question...
Click to expand...
I can't answer it either...
Seems quite orderly; first few, in ascending order:
Code: 
  a    b    c   a+c
  3    2    1     4
  5    6    4     9
  7   12    9    16 
  9   20   16    25
 11   30   25    36
 13   42   36    49
...and so on, a being odd
So next one will be: a=15, b=56, c=49, a+c = 64

And, as you can see, not only c is a square; a+c also is :shock:
 
Here's the site where I found the problem:

Hence c must be a square. For if c and a+c have a common factor, then so do a and c and hence a, b and c, which means they cannot be the minimal set.
Click to expand...

Go [url =http://www.kalva.demon.co.uk/usa/usoln/usol911.html]here [/url] if you want to look at the original problem and the rest of the solution.

Also, what is wrong with my syntax above?
 
Do not put the space after the url:

here

Notice [url=http.....
Not [url= http.........

Quote to see the difference.

Got to be very careful with little things like that when useing Tex and other BBCODES.
 
His solution triangle 16-28-33 is certainly correct: angles ~29, ~58 and ~93.
What I find interesting is the one I show in my list 9-16-20 has obtuse angle
equal to twice another angle: ~26, ~51.5, ~103.

The way he says it:
"Now the triangle with smallest perimeter will have a, b, c relatively prime (otherwise we could divide by the common factor). Hence c must be a square."

sure makes it sound as if it is some theorem we should all know :shock:
 
I now think the answer is as simple as this:
b^2 = c(a + c)
so: b = sqrt(c) * sqrt(a + c)
since b is an integer, then c must be a square, and so must (a + c)
...why in heck didn't we see that sooner :shock:


Btw, there's 2 typoes in your buddy John Scholes' solution:

"...and c^2 = a^2 + b^2 - 2ab sin C" : should be 2ab cos C

"Hence, b / 2c = sin 2C / sin C" : should be sin 2C / 2sin C
 
You must log in or register to reply here.
Top