# 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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er3z7 months agoyes one thing i want to ask you

what

Max Haibara7 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 Mathematics7 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.

I don't understand your problem.

Max Haibara7 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

triton626747 months agoThe theorem he uses supposes it's already true for any integer and he then uses a specific one

Easy Mathematics7 months ago^{+1}If it is valid for any then it is valid for specific.

adandap7 months ago^{+5}That is a sophisticated and elegant argument. I really have to stop just charging ahead with algebra!

Sudheer Thunga7 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 Thunga7 months agoOh... thank you

LetsSolveMathProblems7 months agoYes, 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 erena7 months agoFirst

{GLD} p0pc0r0n7 months ago^{+10}I wanna see letâ€™s solve math problems vs blackpenredpen