A viewer suggested algebra problem.

[@robertgerbicz]

Faster way: from (a+6)^2=4*a*b a is clearly even, and divides (a+6)^2, hence a|36. This leaves only a=2,4,6,12,18,36. We can check easily these, or using that 4 does not divide a we get that a=2,6,18 and all of these are the solutions.

[@fansuli427]

yes

[@f5673-t1h]

To add more words to your solution: a|36 because reducing the equation mod a gives 6^2 is congruent to 0 mod a.

[@gaeb-hd4lf]

Nice 👌🏽

[@Tiqerboy]

I did end up getting a = 2, 6 and 18 and I saw that nothing else would work, but I needed a calculator to work out b in all three cases. If you do that you get b= 14, 12 and 14. So the three solutions (a,b) are (2,14), (6,12) and (18,14) I'm guessing for these kinds of problems, a calculator is not allowed.

[@ricardocavalcanti3343]

@@Tiqerboy You should buy a new calculator. :-) The solutions are (a,b) = (2,8), (6,6) and (18,8).

[@ConradAdamsMrJUMBO]

Wow, what a nice surprise! I've never thought that something I suggested would get featured in a video.🤓 thanks Michael...

[@6d16chanyaukievanzkenith4]

Nicee

[@zeravam]

Congrats! I'm still waiting the exercise I suggested to Michael. Patience is a virtue I have

[@tomatrix7525]

Nice problem

[@loglnlg]

When approximately did you suggest this problem? I'm just asking to have a glimpse idea about time of considering viewers problems

[@adamswilliam3462]

@@loglnlgI think it took at least a few (3-4) weeks!😂 and yes, I have three accounts. 📱🖥📲

[@goodplacetostop2973]

12:53 Homework : Solve this problem over all integers (Z) 13:02 Good Place To Stop

[@WillCummingsvideos]

11:08

[@derekcresswell7352]

These are exactly the type of videos I want to watch. I really enjoy your style. Please keep them coming.

[@raystinger6261]

About the claim 4∤a: Professor Penn tends to make claims ahead of time and then prove them, leaving the student to believe they should had come up with that claim by themselves. But there's a way to naturally come up with that claim. So at 3:28, we came to the conclusion that a must be even. So we can say a=2c, which gives us: (a + 6)² = 4ab => (2c + 6)² = 8bc 4(c + 3)² = 8bc c² + 6c + 9 = 2bc We have that 2bc is even, so the left hand side must also be even. We know 6c is even and 9 is odd, so (6c + 9) is odd, but given that LHS must be even, then c² must be odd as well => c is odd => 4∤a

[@lunstee]

Rather than expand out the square, I would have just left the last line as (c+3)²=2bc . The right side being even implies c+3 is even, which means c must be odd. On the other hand, expanding things out does help with seeing the other constraints. If we divide out your last row by c, we get c+6+9/c = 2b. The left side can only be integer (2b) if c is 1, 3 or 9. All these are solutions.

[@paulbarbat1926]

I think you can take the puzzle a step further : Find all (a, b) among the integers, such that for a given c among the real numbers, you have : Y1 = a*x + c, and Y2 = x*Y1 + b have a unique intersection point

[@SamEden]

"Two times three has a little bit nicer of a form" 😂

[@juanpablogonzalez7130]

If you use quadratic equation many times, you can find that b=6 and b=8 are the only solutions. The case of b=6 is trivial, and the b=8 a consequence from pythagorean triples.

[@donaldbiden7927]

Hi Michael ! Great problem ! Just a way to reduce all the work. When u got (a+6)^2=4ab, u could have just written a^2+a(12-4b)+36=0 \implies the discriminant must be a perfect square. Hence we get b^2-6b=s^2 for some ‘s’. Next, we complete the square by adding a 9 and then factorise things using a^2-b^2=(a+b)(a-b)

[@kevinmartin7760]

I went that way too, but I just solved b(b-6)=s^2 by trial and error. I don't know if I would have to go to the same amount of trouble as Michael did to prove that b=8 (thus a=2 or 18) and b=6 (thus a=6) are the only solutions. You also get b=0 as a solution to b(b-6)=s^2 but this yields a=-6 which is not a natural number and so is not a solution to the overall problem.

[@bilalabbad7954]

I love your explanation .it is clear and organized

[@txikitofandango]

My method can be easily expanded to cover the integers. Set 12-4b to equal the other sums of pairs of factors of 36 (the even ones, obviously). 12-4b = 20 => b = -2 => a = -18 or a = -2 12-4b = 12 => b = 0 => a = -6

[@khaido07]

I love how it began as an algebra problem but then it gets more onto arithmetic when he starts to talk about even numbers and divisibility

[@manucitomx]

This was a great problem. Thank you, professor!

[@renatsamur9099]

Great video. I came up with a slightly different solution for those interested. When we got to (a^2 + 12a +36)/4a = b, I distributed the 1/4a to give a/4 +3 + 9/a which as b is a natural number, a/4 + 9/a is also a natural number. This gives (a^2 +36)/4a is an integer so 4a|a^2+36. This means 4|a^2 +36 and a|a^2+36 giving 4|a^2 and a|36. From 4|a^2, a is even. Going back to a/4 + 9/a is a natural number we can see that as a is even, 9/a is not a natural number so a/4 cannot be one too so 4 does not divide a. Now choosing all a which are even, not divisible by 4 and factors of 36 gives a=2,6,18 with solutions b=8,6,8 respectively from (a^2 + 12a +36)/4a = b.

[@alric8]

I did this problem exactly the same until 2:48 at which point I divided the entire equation by a. This means that you can just check the 6 values of a which divide 36, which is admittedly boring but far more straightforwards.

[@dneary]

I got there a little more quickly by dividing the equation (a+6)^2=4ab by a, giving 4b=a+12+(36/a) with a = 2 mod 4 - leaving only 3 options that have 4b being an integer: 2, 6, 18 (similarly, over integers, a \in {-18, -6, -2, 2, 6, 18}, b \in {-2, 0, -2, 8, 6, 8})

[@roberttelarket4934]

A good place to START your suggested homework assignment.

[@eneapane7594]

Hello. Could you please switch this cool shade with the old one? I find it really hard concentrating with this color.

[@IamBATMAN13]

Another really great problem, thanks for everything prof

[@lmattas]

Such a Nice resolve to the problem, thanks!!!

[@mpdcm]

You also have to check the case a=0

[@mikeh283]

(a+6-2b)^2-(2b-6)^2=6^2. By the Diff of two squares: a-4b+12=(one divisor of 36), a= (the other, corresponding divisor of 36). How many ways are there to split 36? Any choice involves 4 solutions with permutations and negatives.

[@liberatednow6013]

Hi could you try out some math in higher dimensions... that would be cool!

[@guigazalu]

You're saying, like, complex numbers and stuff?

[@aaademed]

When you think that it's an algebraic problem but most of the time we spend on number theory

[@arjunverma963]

7:42 could someone explain what he meant by it, i wasnt able to grasp it

[@DavidCorneth]

He writes a as twice an odd number, i.e. {2, 6, 10, 14, ...}. However he already checked 2 so he's left with a in {6, 10, 14, ...}. Any of those numbers is divisible by an odd prime p. He goes to show that this odd prime must be 3. Makes more sense?

[@stewartcopeland4950]

@@DavidCorneth Why do we exclude for example the value a = 2x3x5 = 30 which also fulfills the 3 conditions: 30 is even, is not a multiple of 4 and 3 divides 30 ?

[@DavidCorneth]

@@stewartcopeland4950 Yes, it fulfills those three conditions but another fails. We have (a + 6)^2 = 4*a*b. If some prime p | a is larger than 3 then we also know that p | 4*a*b since p | a and a | 4*a*b. However p doesn't divide a + 6 since if p | a + 6 and p | a then by doing one step of the euclidean algorithm we have p | 6. But we said p > 3 so that's impossible. In particular 5 doesn't divide 6 so we can immediately see that a = 30 can't give a solution since 5 | 30 and 5 is a prime > 3. See?

[@DavidCorneth]

@@stewartcopeland4950 Maybe try this: find a such that 5 | a and 5 | a + 6. Can you find such a?

[@stewartcopeland4950]

@@DavidCorneth Thanks for your lighting, it's nice to see that you are concerned about "after sales service"

[@debdami]

Does N include 0 ? In that case, any b will fit for a=0. No parabola, but two lines, one horizontal and one of slope 6, with only one intersection.

[@shahjahanishaq1728]

How do we come up with the claim that 'a' cannot be a multiple of 4 in the first place? What does that 'playing around a little bit' refer to?

[@hazalouldi7130]

what about 2 solutions reals or complexes?there are many integers

[@wospy1091]

I did it slightly different. I got -36=a(a+12-4b), so a is a divisor of 36 (1,2,3,4,6,9,12, or 36). Then it's just a matter of checking those 8 cases.

[@mcwulf25]

Excellent, more than one solution! How did he know to check a as a multiple of 4?

[@f5673-t1h]

If the natural numbers are defined to contain 0, then a = 0 and b any natural are solutions.

[@adamswilliam3462]

“a” can’t be 0 since y = ax^2 + 6x + b wouldn’t be a parabola.

[@henkhu100]

@@adamswilliam3462 Why should the graf be a parabola? That is not mentioned in the problem definition

[@williamadams137]

@@henkhu100 At the very start Michael said it was a parabola. Furthermore, it was how I wrote the problem. But sure, if the problem statement was literally “Find all pairs of integers (a,b) such that the graphs of y = ax^2 + 6x + b and y = ax + 6 intersect exactly once.” , then a = 0 and b being any integer would be solutions.

[@henkhu100]

​@@adamswilliam3462 If you look at the problem definition you don't see anything about the fact that the graph of the function has to be a parabola. That Michael makes that statement is his own interpretation of the problem, In fact he solves another problem than the problem on the board.

[@henkhu100]

@@williamadams137 If you look at the problem definition you don't see anything about the fact that the graph of the function has to be a parabola. That Michael makes that statement is his own interpretation of the problem, In fact he solves another problem than the problem on the board. And my remark is about that problem because that's the problem we have to solve.

[@knotwilg3596]

There are a few ways to get to the fact that (a+6)²/4a = (a²+12a+36)/4a must be natural from that follows that 4|a² so a=2k and 2k|k²+6k+9 so k|9 so k=1,3,9 so a=2,6,18

[@DavidCorneth]

ax^2 + 6x + b = ax + 6 gives (ax + 6)*(1-x) = b. The parabola has zeros at x = 1 or x = -6/a so the vertex at x = (1 + -6/a)/2 = 1/2 - 3/a. Plugging that in gives after some shuffling (a + 6)^2 = 4ab. odds and multiples of 4 give a contradiction so a is of the form 4k + 2. Furthermore we have 36 = a*(4b - a - 12) so a|36. The odd divisors of 36 are 1, 3 and 9 so possible values of a are twice those i.e. a in {2, 6, 18} checking those gives the desired solutions (a, b) in {(2, 8), (6, 6), (18, 8)} as shown by Michael.

[@tanusrimalakar9357]

Superb explaination Sir

[@clembat1]

Where can we suggest problems please? Thanks!

[@alfykerolous1869]

why michael all these stuff we observe that (a+6)^2 is a perfect then 4 ab must be a perfect too so a =b and so on a =6 and by ur observatoin 4ab is even then a+6 is even then a at least =2 at the form of (2n) set it = 2n and solve u found that 2b= n+6+9/n and bcs b is and int then n must be equal to 1 2 3 9 and by trying all of them we find the correct one s

[@Bozzigmupp]

At 8:45, why does P have to be able to be divisible by 6

[@DavidCorneth]

We have p | a and p | a+6 so by applying one step of the euclidean algorithm we have p | (a + 6) - a = 6 i.e. p | 6.

[@2070user]

Another way is to rewrite those into equations. p|a and p|(a+6) means a=p*k₁ and (a+6)=p*k₂ where k₁ and k₂ are natural numbers we can substitute a=pk₁ into the second equation pk₁+6=pk₂ 6=pk₂-pk₁=p(k₂-k₁) since 6=p*(k₂-k₁) so that means p|6

[@txikitofandango]

I did it in a totally different way, so it was fun to see a different approach than what I used, and how they intersect! The equation ax^2 + 6x + b = ax + 6 has exactly one solution in x. Equivalently: ax^2 + (6-a)x + b-6 = 0 has one solution. The discriminant (6-a)^2 - 4a(b-6) = 0 a^2 + (12-4b)a + 36 = 0 This quadratic in a will factor when 12 - 4b = plus or minus 12, 13, 15, 20, or 37, which is what you get by multiplying pairs of factors of 36. The only natural numbers b that satisfy this are 6 and 8. The middle coefficient of the quadratic can be 12 - 4(6) = -12 or 12 - 4(8) = -20. If b = 6 then a^2 - 12a + 36 = 0 => a = 6 If b = 8 then a^2 - 20a + 36 = 0 => a = 2 or a = 18 Solutions: (6,6), (2,8), (18,8) Let's see how I did!

[@txikitofandango]

Cool, I got the right answers in a totally different way. So I learned double!

[@emmepombar3328]

The question is also: Find all tangent lines g for parabola p. Couldn't we use, that a is just p'?

[@hasanjakir360]

What if a=0 ?

[@tonyhaddad1394]

(a,b) (2,8) (6,6) (18,8)

[@Happy_Abe]

How by showing p=3 and knowing p divides a do we know a has to be 2(3^m) Why can’t it be two times 3 times some odd number because p=3 divides 3 times an odd.

[@petersievert6830]

Because that would mean there is another odd prime factor in a. But any prime factor of a needs to divide 6. There is no other odd prime factor but 3 doing that.

[@Happy_Abe]

@@petersievert6830 thanks

[@siulibasak3804]

We can also solve it by doing discriminant = 0 (6-a)²-4a(b-6)=0 => a²-(4b-12)a+36=0 mn=36 & m+n=4b-12 After solving this we get all the solutions.....

[@prathmeshraut1616]

Micheal I am beginning my calculas of Higher education any suggestion u would like to give

[@horishok_voloskiy]

те відчуття, коли в старших класах звичайної загальноосвітньої школі строчив тригонометрію як соняшникове насіння, а зараз, через майже 20 років ледве згадуєш ті формули і рівняння

[@ricardocavalcanti3343]

Não entendi.

[@cernejr]

Very nice.

[@karl131058]

Why are people so fond of proof by contradiction? To prove 4 does not divide a, use: a IS even, so a =: 2d => (2d+6)^2=4*2db ! Now pull out 2^2 from the square on the left side and divide by 4 => (d+3)^2=2db ! Right side is even, so left side is even, too. So d+3 is even, so d is odd! (done)

[@CM63_France]

Hi, Ok, great!

[@hitchikerspie]

λ=(z+i)/(z-i) Where z = a + ib When does λ belong to the real numbers?

[@tobiasgorgen7592]

b = a³ / (a²+a) ?

[@manjunathbhat6934]

When a = 0 and b not equal to 1

[@hitchikerspie]

@Federico Rulli Yeah this is it, but the proof is a bit longer expanding out and multiplying by complex conjugates :)

[@tobiasgorgen7592]

@@hitchikerspie hm, I wonder where I went wrong. I expanded the Frac and ended up with a complex number where leting "b" = 0 left me with my solution.

[@36sufchan]

If λ is real, so is λ+1/λ-1 using good ol componendo dividendo, z/i is real => z is purely imaginary Though I don't know how I'll show this comprises of all the solutions

[@Tiessie]

This color palette feels a bit too cold

[@giuseppemalaguti435]

Come sei caduto in basso!!!!

[@haziqthebiohazard3661]

a = 0 and b is any number works too doesn't it

[@TJStellmach]

Not over the natural numbers (given the meaning of that term specified in the video).

[@mepoor761]

I love this color

[@Dshado]

You said you were playing with colour palettes, and I don't think this one is nice. It's too blue and "surgical" in my opinion

[@呂永志]

missing solution: a=0, b is any natural number.

[@tranbachnguyen5108]

Not a good place to stop yet. You forgot a=0.

[@TJStellmach]

No, he didn't. Maybe re-watch the video to see why.

[@armacham]

There are two "a" and two "b" in this problem. Not good! The first "a" is listed in the problem itself (y=ax+6). The second "a" is from the quadratic formula. The same exact thing happens with the letter "b."

[@advaykumar9726]

First