MHB Proving $(a+b,\ a^2-ab+b^2)=1$ or $3$

  • Thread starter Thread starter alexmahone
  • Start date Start date
AI Thread Summary
If \( (a, b) = 1 \), then \( (a+b, a^2-ab+b^2) \) must equal either 1 or 3. The expression simplifies to \( (a+b, 3ab) \), leading to two cases based on the divisibility of \( a+b \) by 3. If \( (a+b, 3) = 1 \), then \( (a+b, 3ab) = 1 \) follows. Conversely, if \( (a+b, 3) = 3 \), it implies that \( 3 \) divides \( a+b \), confirming the result. The proof effectively shows that the greatest common divisor can only be 1 or 3 under the given conditions.
alexmahone
Messages
303
Reaction score
0
Show that if $(a, b)=1$ then $(a+b,\ a^2-ab+b^2)=1\text{ or }3$.

My attempt:

$(a+b,\ a^2-ab+b^2)=(a+b,\ a^2-ab+b^2-(a+b)(a+b))=(a+b,\ -3ab)=(a+b,\ 3ab)$

$(a,\ a+b)=(b,\ a+b)=1$

$\therefore (a+b,\ ab)=1$

Consider 2 cases:

1) $(a+b,\ 3)=1 \implies(a+b,\ 3ab)=1$

2) $(a+b,\ 3)=3\implies 3|a+b$

$\displaystyle (a+b,\ ab)=1\implies\left(\frac{a+b}{3},\ ab\right)=1$

$\displaystyle (a+b,\ 3ab)=3\left(\frac{a+b}{3},\ ab\right)=3$

Is that ok?
 
Last edited:
Mathematics news on Phys.org
Alexmahone said:
Show that if $(a, b)=1$ then $(a+b,\ a^2-ab+b^2)=1\text{ or }3$.

My attempt:

$(a+b,\ a^2-ab+b^2)=(a+b,\ a^2-ab+b^2-(a+b)(a+b))=(a+b,\ -3ab)=(a+b,\ 3ab)$

$(a,\ a+b)=(b,\ a+b)=1$

$\therefore (a+b,\ ab)=1$

Consider 2 cases:

1) $(a+b,\ 3)=1 \implies(a+b,\ 3ab)=1$

2) $(a+b,\ 3)=3\implies 3|a+b$

$\displaystyle (a+b,\ ab)=1\implies\left(\frac{a+b}{3},\ ab\right)=1$

$\displaystyle (a+b,\ 3ab)=3\left(\frac{a+b}{3},\ ab\right)=3$

Is that ok?
Yes. Perfect.
 
You might be interested in the following approach:

Let $d$ be a common divisor of $a+b$ and $a^2-ab+b^2$. Then $ d|(a+b)$ and this can be rewritten as $ a \equiv - b (\bmod. d)$

And so $0\equiv a^2-ab+b^2 \equiv (-b)^2 - (-b) \cdot b + b^2 = 3 b^2 (\bmod. d)$ . But we must have $(d,b) = 1$ since $(a,b)=1$ and $d | (a+b)$ (because any common divisor of $b$ and $d$ will also divide $a$).

Thus $0 \equiv 3 (\bmod. d)$ ...
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top