Number Theory GCD question

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.
 
Jomo said:
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.
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But it's always true that [imath]h^2 | h^2k_1k_2[/imath], is not it?
 
blamocur said:
But it's always true that [imath]h^2 | h^2k_1k_2[/imath], is not it?
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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??
 
Jomo said:
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??
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No argument about the implication.
 
blamocur said:
Nice! It looks to me that you are one step from the proof.
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Could you help me out with that one step because I do not have idea how to proceed further?

Jomo said:
Where did you show that GCD(a, b)< sqrt(a+b)??????
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Yeah that is what I am asking for. what should I do next so that I could complete my proof.

Somebody please help!!
 
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 said:
[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.
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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.
 
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