Can anyone please review/verify my proofs for gcd problem?

[Math100]

Homework Statement:

Assuming that gcd(a, b)=1, prove the following:
(a) gcd(a+b, a-b)=1 or 2.
[Hint: Let d=gcd(a+b, a-b) and show that d$\mid$ 2a, d$\mid$2b, and thus that d$\leq$gcd(2a, 2b)=2 gcd(a, b).]
(b) gcd(2a+b, a+2b)=1 or 3.
(c) gcd(a+b, a^{2}+b^{2})=1 or 2.
[Hint: a^{2}+b^{2}=(a+b)(a-b)+2b^{2}.]
(d) gcd(a+b, a^{2}-ab+b^{2})=1 or 3.
[Hint: a^{2}-ab+b^{2}=(a+b)^{2}-3ab.]

Relevant Equations:

None.

Proof: (a) Suppose that gcd(a, b)=1.
Let d=gcd(a+b, a-b).
By definition of the greatest common divisor,
we have that d$\mid$(a+b) and d$\mid$(a-b).
This means d$\mid$[(a+b)+(a-b)] and d$\mid$[(a+b)-(a-b)],
so we have d$\mid$2a and d$\mid$2b.
Note that d$\leq$gcd(2a, 2b)=2 gcd(a, b).
Since gcd(a, b)=1, it follows that 2 gcd(a, b)=2(1)=2.
Thus, d$\mid$2, which implies that d=1 or d=2.
Therefore, gcd(a+b, a-b)=1 or 2.
(b) Suppose that gcd(a, b)=1.
Let d=gcd(2a+b, a+2b).
By definition of the greatest common divisor,
we have that d$\mid$(2a+b) and d$\mid$(a+2b).
This means d$\mid$[2(2a+b)-(a+2b)] and d$\mid$[-(2a+b)+2(a+2b)],
so we have d$\mid$3a and d$\mid$3b.
Note that d$\leq$gcd(3a, 3b)=3 gcd(a, b).
Since gcd(a, b)=1, it follows that 3 gcd(a, b)=3(1)=3.
Thus, d$\mid$3, which implies that d=1 or d=3.
Therefore, gcd(2a+b, a+2b)=1 or 3.
(c) Suppose that gcd(a, b)=1.
Let d=gcd(a+b, a^{2}+b^{2}).
By definition of the greatest common divisor,
we have that d$\mid$(a+b) and d$\mid$(a^{2}+b^{2}).
This means d$\mid$[(a^{2}+b^{2})+(a^{2}-b^{2})] and d$\mid$[(a^{2}+b^{2})-(a^{2}-b^{2})],
so we have d$\mid$2a^{2} and d$\mid$2b^{2}.
Note that d$\leq$gcd(2a^{2}, 2b^{2})=2 gcd(a^{2}, b^{2}).
Since gcd(a, b)=1, it follows that gcd(a^{2}, b^{2})=1.
Thus, 2 gcd(a^{2}, b^{2})=2(1)=2, and so d$\mid$2,
which implies that d=1 or d=2.
Therefore, gcd(a+b, a^{2}+b^{2})=1 or 2.
(d) Suppose that gcd(a, b)=1.
Let d=gcd(a+b, a^{2}-ab+b^{2}).
By definition of the greatest common divisor,
we have that d$\mid$(a+b) and d$\mid$(a^{2}-ab+b^{2}).
This means d$\mid$[(a+b)^{2}-(a^{2}-ab+b^{2})], so we have d$\mid$3ab.
Note that each prime divisor k of d must divide either 3, a or b.
Thus, we have that d$\mid$[3a(a+b)-3ab] and d$\mid$[3b(a+b)-3ab],
so d$\mid$3a^{2} and d$\mid$3b^{2}.
Then we get d$\leq$gcd(3a^{2}, 3b^{2})=3 gcd(a^{2}, b^{2}).
Since gcd(a, b)=1, it follows that gcd(a^{2}, b^{2})=1.
Thus, 3 gcd(a^{2}, b^{2})=3(1)=3, and so d$\mid$3,
which implies that d=1 or d=3.
Therefore, gcd(a+b, a^{2}-ab+b^{2})=1 or 3.

 

[fishturtle1]

Homework Statement:: Assuming that gcd(a, b)=1, prove the following:
(a) gcd(a+b, a-b)=1 or 2.
[Hint: Let d=gcd(a+b, a-b) and show that d$\mid$ 2a, d$\mid$2b, and thus that d$\leq$gcd(2a, 2b)=2 gcd(a, b).]
(b) gcd(2a+b, a+2b)=1 or 3.
(c) gcd(a+b, a^{2}+b^{2})=1 or 2.
[Hint: a^{2}+b^{2}=(a+b)(a-b)+2b^{2}.]
(d) gcd(a+b, a^{2}-ab+b^{2})=1 or 3.
[Hint: a^{2}-ab+b^{2}=(a+b)^{2}-3ab.]
Relevant Equations:: None.

Note that d≤gcd(3a, 3b)=3 gcd(a, b).
Since gcd(a, b)=1, it follows that 3 gcd(a, b)=3(1)=3.
Thus, d∣3, which implies that d=1 or d=3.

b) From the first and second line, we can conclude $d \le 3$. But that doesn't imply $d$ divides $3$. But this can be fixed by changing the first line to "Note that $d$ divides $\gcd(3a,3b) = 3\gcd(a, b)$". You may want to make a similar change in a, c, d) even though it kind of works out in a) and c) case.

Homework Statement:: Assuming that gcd(a, b)=1, prove the following:
(a) gcd(a+b, a-b)=1 or 2.
[Hint: Let d=gcd(a+b, a-b) and show that d$\mid$ 2a, d$\mid$2b, and thus that d$\leq$gcd(2a, 2b)=2 gcd(a, b).]
(b) gcd(2a+b, a+2b)=1 or 3.
(c) gcd(a+b, a^{2}+b^{2})=1 or 2.
[Hint: a^{2}+b^{2}=(a+b)(a-b)+2b^{2}.]
(d) gcd(a+b, a^{2}-ab+b^{2})=1 or 3.
[Hint: a^{2}-ab+b^{2}=(a+b)^{2}-3ab.]
Relevant Equations:: None.

By definition of the greatest common divisor,
we have that d∣(a+b) and d∣(a^{2}+b^{2}).
This means d∣[(a^{2}+b^{2})+(a^{2}-b^{2})] and d∣[(a^{2}+b^{2})-(a^{2}-b^{2})]

c) Maybe I'm just being slow, but i think it would be helpful to include a line "Since $d \vert (a+b)$, we have $d \vert (a+b)(a-b) = a^2 - b^2$. Other than that, i think everything else is correct.

Homework Statement:: Assuming that gcd(a, b)=1, prove the following:
(a) gcd(a+b, a-b)=1 or 2.
[Hint: Let d=gcd(a+b, a-b) and show that d$\mid$ 2a, d$\mid$2b, and thus that d$\leq$gcd(2a, 2b)=2 gcd(a, b).]
(b) gcd(2a+b, a+2b)=1 or 3.
(c) gcd(a+b, a^{2}+b^{2})=1 or 2.
[Hint: a^{2}+b^{2}=(a+b)(a-b)+2b^{2}.]
(d) gcd(a+b, a^{2}-ab+b^{2})=1 or 3.
[Hint: a^{2}-ab+b^{2}=(a+b)^{2}-3ab.]
Relevant Equations:: None.

Note that each prime divisor k of d must divide either 3, a or b.
Thus, we have that d∣[3a(a+b)-3ab] and d∣[3b(a+b)-3ab]

d) I don't think the first line implies the second, i.e., if a prime $p$ divides $b$ but not $a$ or $3$, then we have $p$ doesn't divide $3a(a+b) - 3ab$. However, the second line is true since $d \vert (a+b)$ and $d \vert 3ab$. Other than that, and changing $d \le 3$ to $d \vert 3$, it looks good to me.

 

[Math100]

I revised this problem, can you please review/verify to see if it's correct this time?

Proof:
(a) Suppose that gcd(a, b)=1.
Let d=gcd(a+b, a-b).
By definition of the greatest common divisor,
we have that d$\mid$(a+b) and d$\mid$(a-b).
This means d$\mid$[(a+b)+(a-b)] and d$\mid$[(a+b)-(a-b)],
so we have d$\mid$2a and d$\mid$2b.
Note that d divides gcd(2a, 2b)=2 gcd(a, b).
Since gcd(a, b)=1, it follows that 2 gcd(a, b)=2(1)=2.
Thus, d$\mid$2, which implies that d=1 or d=2.
Therefore, gcd(a+b, a-b)=1 or 2.
(b) Suppose that gcd(a, b)=1.
Let d=gcd(2a+b, a+2b).
By definition of the greatest common divisor,
we have that d$\mid$(2a+b) and d$\mid$(a+2b).
This means d$\mid$[2(2a+b)-(a+2b)] and d$\mid$[-(2a+b)+2(a+2b)],
so we have d$\mid$3a and d$\mid$3b.
Note that d divides gcd(3a, 3b)=3 gcd(a, b).
Since gcd(a, b)=1, it follows that 3 gcd(a, b)=3(1)=3.
Thus, d$\mid$3, which implies that d=1 or d=3.
Therefore, gcd(2a+b, a+2b)=1 or 3.
(c) Suppose that gcd(a, b)=1.
Let d=gcd(a+b, a^{2}+b^{2}).
By definition of the greatest common divisor,
we have that d$\mid$(a+b) and d$\mid$(a^{2}+b^{2}).
Since d$\mid$(a+b), we have d$\mid$(a+b)(a-b)=a^{2}-b^{2}.
This means d$\mid$[(a^{2}+b^{2})+(a^{2}-b^{2})] and d$\mid$[(a^{2}+b^{2})-(a^{2}-b^{2})],
so we have d$\mid$2a^{2} and d$\mid$2b^{2}.
Note that d divides gcd(2a^{2}, 2b^{2})=2 gcd(a^{2}, b^{2}).
Since gcd(a, b)=1, it follows that gcd(a^{2}, b^{2})=1.
Thus, 2 gcd(a^{2}, b^{2})=2(1)=2, and so d$\mid$2,
which implies that d=1 or d=2.
Therefore, gcd(a+b, a^{2}+b^{2})=1 or 2.
(d) Suppose that gcd(a, b)=1.
Let d=gcd(a+b, a^{2}-ab+b^{2}).
By definition of the greatest common divisor,
we have that d$\mid$(a+b) and d$\mid$(a^{2}-ab+b^{2}).
This means d$\mid$[(a+b)^{2}-(a^{2}-ab+b^{2})],
so we have d$\mid$3ab.
Now we have that d$\mid$[3a(a+b)-3ab] and d$\mid$[3b(a+b)-3ab],
so d$\mid$3a^{2} and d$\mid$3b^{2}.
Note that d divides gcd(3a^{2}, 3b^{2})=3 gcd(a^{2}, b^{2}).
Since gcd(a, b)=1, it follows that gcd(a^{2}, b^{2})=1.
Thus, 3 gcd(a^{2}, b^{2})=3(1)=3, and so d$\mid$3,
which implies that d=1 or d=3.
Therefore, gcd(a+b, a^{2}-ab+b^{2})=1 or 3.

I am so sorry for the late response of this problem. I was too busy in my workplace for the past two days. Now this problem is revised. Please leave any comment or feedback. I want to know if it's good or not.

 

[fishturtle1]

I revised this problem, can you please review/verify to see if it's correct this time?

Proof:
(a) Suppose that gcd(a, b)=1.
Let d=gcd(a+b, a-b).
By definition of the greatest common divisor,
we have that d$\mid$(a+b) and d$\mid$(a-b).
This means d$\mid$[(a+b)+(a-b)] and d$\mid$[(a+b)-(a-b)],
so we have d$\mid$2a and d$\mid$2b.
Note that d divides gcd(2a, 2b)=2 gcd(a, b).
Since gcd(a, b)=1, it follows that 2 gcd(a, b)=2(1)=2.
Thus, d$\mid$2, which implies that d=1 or d=2.
Therefore, gcd(a+b, a-b)=1 or 2.
(b) Suppose that gcd(a, b)=1.
Let d=gcd(2a+b, a+2b).
By definition of the greatest common divisor,
we have that d$\mid$(2a+b) and d$\mid$(a+2b).
This means d$\mid$[2(2a+b)-(a+2b)] and d$\mid$[-(2a+b)+2(a+2b)],
so we have d$\mid$3a and d$\mid$3b.
Note that d divides gcd(3a, 3b)=3 gcd(a, b).
Since gcd(a, b)=1, it follows that 3 gcd(a, b)=3(1)=3.
Thus, d$\mid$3, which implies that d=1 or d=3.
Therefore, gcd(2a+b, a+2b)=1 or 3.
(c) Suppose that gcd(a, b)=1.
Let d=gcd(a+b, a^{2}+b^{2}).
By definition of the greatest common divisor,
we have that d$\mid$(a+b) and d$\mid$(a^{2}+b^{2}).
Since d$\mid$(a+b), we have d$\mid$(a+b)(a-b)=a^{2}-b^{2}.
This means d$\mid$[(a^{2}+b^{2})+(a^{2}-b^{2})] and d$\mid$[(a^{2}+b^{2})-(a^{2}-b^{2})],
so we have d$\mid$2a^{2} and d$\mid$2b^{2}.
Note that d divides gcd(2a^{2}, 2b^{2})=2 gcd(a^{2}, b^{2}).
Since gcd(a, b)=1, it follows that gcd(a^{2}, b^{2})=1.
Thus, 2 gcd(a^{2}, b^{2})=2(1)=2, and so d$\mid$2,
which implies that d=1 or d=2.
Therefore, gcd(a+b, a^{2}+b^{2})=1 or 2.
(d) Suppose that gcd(a, b)=1.
Let d=gcd(a+b, a^{2}-ab+b^{2}).
By definition of the greatest common divisor,
we have that d$\mid$(a+b) and d$\mid$(a^{2}-ab+b^{2}).
This means d$\mid$[(a+b)^{2}-(a^{2}-ab+b^{2})],
so we have d$\mid$3ab.
Now we have that d$\mid$[3a(a+b)-3ab] and d$\mid$[3b(a+b)-3ab],
so d$\mid$3a^{2} and d$\mid$3b^{2}.
Note that d divides gcd(3a^{2}, 3b^{2})=3 gcd(a^{2}, b^{2}).
Since gcd(a, b)=1, it follows that gcd(a^{2}, b^{2})=1.
Thus, 3 gcd(a^{2}, b^{2})=3(1)=3, and so d$\mid$3,
which implies that d=1 or d=3.
Therefore, gcd(a+b, a^{2}-ab+b^{2})=1 or 3.

I am so sorry for the late response of this problem. I was too busy in my workplace for the past two days. Now this problem is revised. Please leave any comment or feedback. I want to know if it's good or not.

No need to apologize; the proofs look correct to me.

 

[Math100]

Thank you so much for the help!