# Solution 76: Polygonal Numbers and GCD's

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• Published on Jan 24, 2019
• Let's use properties of divisors to gain information about b_n.
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• er3z 7 months ago

yes one thing i want to ask you

what

• Max Haibara 7 months ago +1

wait... you said that if b|a and b|c, then b|ax+cy for ANY integer x,y. But then you only use SPECIFIC value of x,y which is n+1 and n-1 to prove it. How about the other value of x,y?

• Easy Mathematics 7 months ago

@Max Haibara The fact he uses is easy to proof. There is no problem.
if a | b and a | c, then a | xb + yc for any x,y.
In words:
If a divides two integers, then a divides every linear combination.
This fact is easy to proof.
So he choosed a specific one. For purpose. That's all and everything is correct.

• Max Haibara 7 months ago

@Easy Mathematics but not all ax+cy will be equal to 1, at least we need to prove that all x,y will make ax+cy=1
If there is a value of x,y that makes ax+cy!=1, then it's not valid

• triton62674 7 months ago

The theorem he uses supposes it's already true for any integer and he then uses a specific one

• Easy Mathematics 7 months ago +1

If it is valid for any then it is valid for specific.

• adandap 7 months ago +5

That is a sophisticated and elegant argument. I really have to stop just charging ahead with algebra!

• Sudheer Thunga 7 months ago +2

So...the theorem you stated at 4:51 states that when c is a divisor of gcd(a,b)... c is also a divisor of a and b.

• Sudheer Thunga 7 months ago

Oh... thank you

• LetsSolveMathProblems  7 months ago

Yes, you are correct. As a fun fact, the converse is true, as well: If c is a divisor of a and b, it is a divisor of gcd(a,b). The converse is slightly harder to prove.

• mark erena 7 months ago

First

• {GLD} p0pc0r0n 7 months ago +10

I wanna see letâ€™s solve math problems vs blackpenredpen