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find all natural numbers
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aadil
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#1
find all natural numbers
find all natural numbers a,b,c such that ab+bc+ca=a+b+c+1
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PostPosted: Sat Nov 07, 2009 4:07 am  Back to top 
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math154
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#2
Solution
Let x=abc so that
(a-1)(b-1)(c-1)=x-(ab+bc+ca-a-b-c)-1=x-2
and
2=abc-(a-1)(b-1)(c-1)>(b-1)(c-1).
If b=1 or c=1, then x-2=0\implies x=abc=2, which yields the permutations of \boxed{(2,1,1)} as solutions.
If b,c\ge2, then if b\ge3 or c\ge3 we have a contradiction, so b=c=2\implies x=4a,\quad x-2=a-1\implies 2=3a+1\implies a=\frac13, which is also a contradiction.

PostPosted: Sat Nov 07, 2009 1:26 pm  Back to top 
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Mateescu Constantin
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#3
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ab+bc+ca=a+b+c+1\ \Longleftrightarrow\ a(b+c)+bc=a+1+b+c

\Longleftrightarrow\ (b+c)(a-1)=a+1-bc\ \Longleftrightarrow\ (b+c)(a-1)=a-1+2-bc

\Longleftrightarrow\ (a-1)(b+c-1)=2-bc\ (1). Because 2-bc\in\mathbb{N}\ \implies bc\in\{0,1,2\}

1) bc=0. Case b=c=0 is false, so let b=0.

Then, from (1) we obtain that (a-1)(c-1)=2, which gives a=2, c=3 or a=3,c=2 .

2) bc=1\ \implies b=c=1 and a=2 .

3) bc=2\ \implies b=1,c=2 or b=2,c=1 and according to (1) we have a=1 .

Therefore the solutions of the equation are (1,1,2);(0,2,3) and their permutations.



PostPosted: Sat Nov 07, 2009 1:44 pm  Back to top 
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math154
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#4
If we are taking natural numbers to include zero, then
Mateescu Constantin wrote:
\Longleftrightarrow\ (a - 1)(b + c - 1) = 2 - bc\ (1). Because 2 - bc\in\mathbb{N}\ \implies bc\in\{0,1,2\}

is flawed because the possibility that a = 0 has not been ruled out. Of course, it's easy to fix.

PostPosted: Sat Nov 07, 2009 1:54 pm  Back to top 
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aadil
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#5
1,1,2 is the only solution in natural numbers
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PostPosted: Sun Nov 08, 2009 1:52 am  Back to top 
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Arrange your tan
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#6
aadil wrote:
1,1,2 is the only solution in natural numbers


(1,1,2)
(1,2,1)
(2,1,1)

These were equivalently referred to already in an earlier post in this thread.

aadil, you should use the phrase "positive integers" and not, unfortunately
the ambiguous phrase "natural numbers," because of its stance of the
inclusion of zero. I always have it as not including zero, and it is
taught that way in Mathematics for Elementary Teachers courses,
for example whenever I come across them.

PostPosted: Sun Nov 08, 2009 3:49 pm  Back to top 
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aadil
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#7
Quote:
aadil, you should use the phrase "positive integers" and not, unfortunately
the ambiguous phrase "natural numbers," because of its stance of the
inclusion of zero. I always have it as not including zero, and it is
taught that way in Mathematics for Elementary Teachers courses,
for example whenever I come across them

0 is definitely not natural.thats where the concept of whole numbers comes into existence
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PostPosted: Mon Nov 09, 2009 6:13 am  Back to top 
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JBL
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#8
That convention is not universal. When I was in elementary school (and maybe high school), we used 0 \not \in {\bf N}, but computer scientists and combinatorialists (I'm the latter) use 0 \in \bf N. The symbol (\bf N or \mathbb{N}) and the name "natural numbers" are both ambiguous, and it would be best to use unambiguous notations like {\bf Z}_{ > 0}, {\bf Z}_{\geq 0}, "positive integers," "nonnegative integers," etc. The phrase "whole numbers" is not used beyond secondary school, as far as I can tell. (There may also be differences between countries or continents about these notations; I'm speaking as an American.)

Is there any k such that there are no solutions to ab + bc + ca = a + b + c + k?
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Joel
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PostPosted: Mon Nov 09, 2009 6:23 am  Back to top 
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digger
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#9
JBL wrote:
Is there any k such that there are no solutions to ab + bc + ca = a + b + c + k?


Take a=b=1, c=k+1 Very Happy

PostPosted: Tue Nov 10, 2009 6:41 pm  Back to top 
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JBL
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#10
Okay, good. There are some values of k for which this is the only solution in positive integers (like k = 1, for example). Can we classify these values of k? (I don't know the answer to this question.)
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Joel
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PostPosted: Tue Nov 10, 2009 6:57 pm  Back to top 
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#11
I know nothing exept trivial cases doubly (at least) presentable k, like

k\equiv 0, 2 \pmod 3
k\equiv 1, 3 \pmod 4
..........
k\equiv i(m + 1 - i) - 1 \pmod m, \ \mbox{where} \ 1\leqslant i \leqslant \frac {m + 1}{2}

for sufficiently large k.

PostPosted: Thu Nov 12, 2009 12:50 am  Back to top 
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