# Solution 76: Polygonal Numbers and GCD's

Embed

**Published on Jan 24, 2019**- Let's use properties of divisors to gain information about b_n.

Congratulations to Minh Cong Nguyen, mark erena, iQuickdraw X, and Xu Chen Tan for successfully solving this math challenge question! Minh Cong Nguyen was the first person to solve the question.

Your support is truly a huge encouragement.

Please take a second to subscribe in order to send us your valuable support and receive notifications for new videos!

Every subscriber and every like are wholeheartedly appreciated.

For more Weekly Math Challenges:

usclip.net/p/PLpoKXj-PWCbaDXYHES37_zX4O-kCWxguM

er3z4 months agoyes one thing i want to ask you

what

Max Haibara4 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 Mathematics4 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 Haibara4 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

triton626744 months agoThe theorem he uses supposes it's already true for any integer and he then uses a specific one

Easy Mathematics4 months ago^{+1}If it is valid for any then it is valid for specific.

adandap4 months ago^{+5}That is a sophisticated and elegant argument. I really have to stop just charging ahead with algebra!

Sudheer Thunga4 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 Thunga4 months agoOh... thank you

LetsSolveMathProblems4 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 erena4 months agoFirst

{GLD} p0pc0r0n4 months ago^{+10}I wanna see let’s solve math problems vs blackpenredpen