Problem for good mathematicians

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leaftrimmer
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#1
Report Thread starter 17 years ago
#1
a and b are positive, coprime integers:

Show that the h.c.f of (a + b) and (a^2 - ab + b^2) is 1 or 3.
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misshn
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#2
Report 17 years ago
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let the hcf of a+b and a^2 -ab + b^2 be k.
=> a+b and a^2-ab+b^2 can divide by k.
=>(a+b)^2 - 3ab can divide by k
=> 3ab can divide by k
1)If k=1 then the statement is right.
2)If k>1 then let p be a prime factor of k.
If p>3 then ab must divide by p. Coz (a,b)=1 then only a or b can divide by p.
=>a+b can not divide by p (contradict with the assumption that p is the hcf)
=>p=3. =>k=3^n(with n>=1 and n is integer)
As a+b can divide by 3 and (a,b)=1 then neither a nor b can divide by 3
=>3ab can divide by 3 but not by 9=>n<2=>n=1 and k=3

So the hcf of a+b and a^2-ab+b^2 can only be 1 or 3
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leaftrimmer
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#3
Report Thread starter 17 years ago
#3
(Original post by misshn)
let the hcf of a+b and a^2 -ab + b^2 be k.
=> a+b and a^2-ab+b^2 can divide by k.
=>(a+b)^2 - 3ab can divide by k
=> 3ab can divide by k
1)If k=1 then the statement is right.
2)If k>1 then let p be a prime factor of k.
If p>3 then ab must divide by p. Coz (a,b)=1 then only a or b can divide by p.
=>a+b can not divide by p (contradict with the assumption that p is the hcf)
=>p=3. =>k=3^n(with n>=1 and n is integer)
As a+b can divide by 3 and (a,b)=1 then neither a nor b can divide by 3
=>3ab can divide by 3 but not by 9=>n<2=>n=1 and k=3

So the hcf of a+b and a^2-ab+b^2 can only be 1 or 3
Well done misshn. I did it in a similar way:

(a+b,a^2-ab+b^2) = (a+b,(a+b)^2-3ab)
a+b obviously divides (a+b)^2
therefore (a+b,(a+b)^2-3ab) = (a+b,3ab)
(a,b) = 1 so (a,ab)=(b,ab)=(a+b,ab)=1
which implies that (a+b,3ab) = 1 or 3

Now try this one, based along similar lines - show that (a+b,(a^p + b^p)/(a+b)) = 1 or p (where p is an odd prime)
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amo1
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#4
Report 17 years ago
#4
damn, obviously not a gd mathmatition
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J.F.N
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#5
Report 17 years ago
#5
(Original post by leaftrimmer)
Now try this one, based along similar lines - show that (a+b,(a^p + b^p)/(a+b)) = 1 or p (where p is an odd prime)
Whats that , doing in the question? Did you mean . ?
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leaftrimmer
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#6
Report Thread starter 17 years ago
#6
(Original post by J.F.N)
Whats that , doing in the question? Did you mean . ?
no i mean , . (x,y) stands for the h.c.f of x and y.
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leaftrimmer
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#7
Report Thread starter 17 years ago
#7
(Original post by amo1)
damn, obviously not a gd mathmatition
What's that supposed to mean?
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