Number Theory GCD question

[AdkAdi]

Please provide with the proper mathematical solution

 

[AdkAdi]

My approach

 

[Steven G]

While it is true that if h|h²k₁k₂, then h²|h²k₁k₂ I would not say that h|h²k₁k₂ implies h²|h²k₁k₂.

That statement means, imo, that if a|b, then a²|b, which is not necessarily true.

 

[Steven G]

View attachment 30797My approach

Where did you show that GCD(a, b)< sqrt(a+b)??????

 

[blamocur]

View attachment 30797My approach

Nice! It looks to me that you are one step from the proof.

 

[blamocur]

While it is true that if h|h²k₁k₂, then h²|h²k₁k₂ I would not say that h|h²k₁k₂ implies h²|h²k₁k₂.

That statement means, imo, that if a|b, then a²|b, which is not necessarily true.

But it's always true that [imath]h^2 | h^2k_1k_2[/imath], is not it?

 

[Steven G]

But it's always true that [imath]h^2 | h^2k_1k_2[/imath], is not it?

Yes, and I did say that. But I still do not accept the fact that h|h₂k₁k₂ implies h²|h²k₁k₂

While 2+3 = 5, I would not say that 1+1=2 implies 2+3=5.
Am I wrong??

 

[blamocur]

Yes, and I did say that. But I still do not accept the fact that h|h₂k₁k₂ implies h²|h²k₁k₂

While 2+3 = 5, I would not say that 1+1=2 implies 2+3=5.
Am I wrong??

No argument about the implication.

 

[Steven G]

No argument about the implication.

OK

 

[AdkAdi]

Nice! It looks to me that you are one step from the proof.

Could you help me out with that one step because I do not have idea how to proceed further?

Where did you show that GCD(a, b)< sqrt(a+b)??????

Yeah that is what I am asking for. what should I do next so that I could complete my proof.

Somebody please help!!

 

[BigBeachBanana]

My approach:
[math]a^2+a+b^2+b=kab\\ a+b=kab-a^2-b^2\\[/math]The RHS is divisible by the [imath]gcd(a,b)^2\medspace[/imath] i.e. [imath]gcd(a,b)^2|a+b \implies a+b=n\cdot gcd(a,b)^2[/imath]. It follows [imath]a+b \ge gcd(a,b)^2[/imath]

 

[blamocur]

Could you help me out with that one step because I do not have idea how to proceed further?

[imath]h^2 | (a+b)(a+b+1)[/imath], but [imath]h^2[/imath] can only divide one of the multipliers because it is mutually prime with another.

 

[BigBeachBanana]

[imath]h^2 | (a+b)(a+b+1)[/imath], but [imath]h^2[/imath] can only divide one of the multipliers because it is mutually prime with another.

To expand on this:
If [imath]h=gcd(a,b)\neq1[/imath], then [imath]k\cdot h+1[/imath] is co-prime with h.
If [imath]h=1[/imath], then the inequality is trivial.
Thus, for [imath]h\neq1[/imath], then [imath]h^2|(a+b)[/imath], which came to the same conclusion I posted in #13.

 

[AdkAdi]

Is this correct? If not then what is the error in this approach?