# Problem for good mathematicians

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leaftrimmer

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#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.

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

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#2

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

=> 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

(Original post by

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

**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

(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

J.F.N

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#5

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#5

(Original post by

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)

**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)

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leaftrimmer

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#6

(Original post by

Whats that , doing in the question? Did you mean . ?

**J.F.N**)Whats that , doing in the question? Did you mean . ?

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#7

(Original post by

damn, obviously not a gd mathmatition

**amo1**)damn, obviously not a gd mathmatition

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