A trick I found to make the solution quicker is that since n² has the same parity as a²b², we can use (ab - 2) as an upper bound for n, which leads to a more restrictive equation than with (ab - 1) and leaves us with only a handful of cases to check.

Hi, For fun: 1 "the next thing that we want to do", 1 "what I want to notice", 1 "so now what we are going to do", 2 "I'll go ahead and", 2 "so let's go ahead and", 1 "so let's may be go ahead and", 1 "I can go ahead and", 1 "we can go ahead and", 1 "and so on and so forth", 1 "great", 6:05 : a good place to look at. 6:31 : 2b or not 2b, that is the question. 19:48 : whouah! whouah !

From a former student of NUIG, Michael Tuite is a legend 😄

ab-2a-2b

One could say he is a Math buff.

I like this kind of problems.. When I watch the solution I feel like "That's easy and funny" but I can't do that alone

There's a small shortcut for the a=2 case. Note that (2b-1)^2=n^2+3^2 is a Pythagorean triple with one side length of 3, implying n=4 and 2b-1=5. Although is does make it harder to realize the 0+9=9 case is actually relevant.

Apparently my first comment did NOT appear for other people so let’s try again... 5:01 Good inequality to stop 6:01 Good place to look at 20:07 Sin áit mhaith le stopadh Daily... How many elements do the following sets have : A = {x ∈ ℚ | x = (n² + 1)/(2n² + n + 1), n ∈ ℕ, n ∈ [1, 1000]} B = {x ∈ ℚ | x = (n² + 1)/(2n² + n + 1), n ∈ ℕ} Posting hint in another comment.

This is the best channel for math on youtube

Man your videos are really nice to watch

Wow! Nice problem! I can’t really say that it was hard, but I have to admit that I enjoyed watching you explain it. Keep up the awesome work and have a good one! P.S. I’m still waiting for that juicy IMO playlist of problems...

Awesome explanation.

The dog was impressed at the end.

Really intuitive and simple solution!

6:00 this can also be thought of as a rectangle with sides a and b, where the area is less than the perimeter! (without regard to units, that is.)

case 1 has also (trivial) subcase 3: ((b-2)-n)((b-2)+n) = 8 may have both outer parenthesis to be negative. But that is possible only for b < 2, thus we should quickly check (and rule out) a=1, b=1.

My thoughts how to solve this: since (ab)^2 is already a square, we need to look at the difference between that and the perfect square just below it. Once a and b get beyond a certain since, this difference is going to be > 4(a+b) and therefore you only really need to test small values of a and b for that to hold. When I solved this , I consider 0 NOT a Natural number. Natural numbers should be 1, 2, 3, .....

The only person who motivates me to become a mathematician!❤️

His videos are awesome 😍 it inspires me to solve olympaids questions

These videos are the best

Fun problem! But the parity argument (as mentioned below) plus some factoring significantly simplifies the cases to check. I also explored a=b in general which leads to the interesting fact that a^4-8a = a*(a-2)*(a^2+2a+4) is square ... but isn't much help in this problem but might be a useful approach for similar problems. (edited)

I found out that for a,b more than or equal to 5, the given expression is between (ab)^2 and (ab-1)^2 , so no solutions there. Then the rest is casework !! Nice problem

I'm not that bright; once figuring out the upper bound I was ready to just plug in everything from 0,0 to 4,4. Then midway through I remembered 0 isn't Natural

I was in Galway University a few weeks ago ... but just as a tourist.

1. if a = 1 we have b^2-4*b-4 it has more than 1 root, thus can't be square but can be equal to 1 when x = 5. [1,5],[5,1]. 2 if a>=b>1 then a*b > a+ b or a+b = a*b-k with k>=1, we're getting x^2 - 4*x + 4*k which is always > 1 but can have one root when k = 1. so a*b = a+b-1 which can only be true in [2,3],[3,2]

Man wowwwww you are great i always learn from you

He's definitely the Bob Ross of math !

Thank you!

Ab+n=1or 2or 4 and ab+n=4(a+b) ,or 2(a+b) Or(a+b) respectively and since the equation is symmetric hence if (a,b) is a solution then (b,a) is also a solution

Im my country, 0 is a natural number So case 3: a=0 -4b=n² As -4b≤0 and n²≥0, b=0

Great sir 😍loved your way

factorize b^2·(a + 2·(√(1 - b^3) + 1)/b^2)·(a - 2·(√(1 - b^3) - 1)/b^2) = n^2 u can see for all values greater than b =1 , we are going to have complex number. if b= 1 then (a+2)^2 = n^2 or a = n -2; if b = 1 then b = n-2; or as u said, a*b

case a=1: If (b-2)^2 = n^2+8 then you have 2 perfect squares which differ by 8. For (16,25) and above there are no solutions. So you look at all couples below that. And it's quickly seen that only (1,9) fits the bill. Similarly for a=2, but this time since the difference is 9 you include(16,25) to be checked which gives a solution.

That a good channel to stop by ......😛😛

I believe for all of the system of equations we should also find n to check that it's integer

Galway when pronounced sounds like "Gaulway" where Gaul sounds like the natives of what is now France.

06:03 a = 2.(a + b); a b = a; Substituting into the inequality:: a² >= 4.a; ==> a >= 4;

I converted this problem to "find two pairs of natural numbers such that the sum of one pair equals the product of the other pair" (see that 1+5=2x3 and 2+3=1x5). My solution is a little longer though. First, draw a square ABCD with side length ab. Then draw the diagonals. Draw a smaller square EFGH inside. the centroid of ABCD and EFGH must coincide. Now let the side length of EFGH be n. Notice that there are 4 trapezoids formed. Let the height of each of the trapezoid be h. Then we can see that n=ab-2h. The task will be to find h such that the area of one trapezoid is (a+b). Hence h(ab-h)=(a+b). We simplify to get the quadratic equation h^2-abh+(a+b)=0. Let's call this equation (1)Notice that the discriminant of this is the problem we are trying to solve. This means that the discriminant is a perfect square >0 and the roots will be real, which are (ab +/- n)/2. If both a and b are odd, then n is odd and so the roots are natural numbers. If either a or b is even, or both are even, then n is also even and h is still a natural number. Hence, we can denote the roots as natural numbers s,t. By inspection of equation (1), we see that st=(a+b) and s+t=ab. We can write another quadratic equation here where a and b are the roots. The quadratic equation is j^2-(st)j+(s+t)=0. Let us call this equation (2). Now we see that the determinant of equation (2) is d=(st)^2-4(s+t), and that both a and b are natural numbers and so this determinant is a perfect square also. This means that is we now a or b, we can use that to find another solution s and t using equation 1 and if we know a solution s and t using equation 2. Now consider first f(x,y)=xy/(x+y). Using calculus, we can see that f(x,y) is increasing for any natural numbers x and y. Notice also that when either x or y=1, then xy/(x+y)=2, y>=2, then f(x,y)>=1. Going back to a,b,s and t, we have st=a+b and s+t=ab (hence find two pairs of natural numbers such that the sum of one pair equals the product of the other pair). From this we have (st)/(s+t)=(a+b)/(ab). We see now that when the LHS >1 then the RHS 1, RHS1, which means that (ab)/(a+b)1, RHS

It would be interesting to see if the original question specifically allowed for n=0

I wonder if including that last solution would have been considered wrong by the Olympiad evaluators, since the problem specified that n was a natural number. At a minimum, I think one would have to qualify that solution with a conditional such as "If n can be 0...).

Man i love math

Applying congruences to the original equation, modulo 4, 6, and 8 might have been useful.

Rather than applying case by case i solved like this. when 4(a+b)=>2ab-1, we can assume that 4(a+b)=>2ab because both terms are even number and we obtain 2a/(a-2)=>b and 2b/(b-2)=>a After that we multiply both sides and cancel out ab terms because it cannot be 0. At the end we can obtain 4=>(a-2)*(b-2).... After than this point we try all combinations.

In the first sight ,I can only get the solution (2.3) .Thanks for the other answer (1,5) .

Keep up broo Can you get us some abstract algebra medium problems .

Quick solution: n^2

Ok, I just want to be sure I'm not misinterpreting this. In this case, you can assume a < or = b because in the original equation, the two terms are interchangeable, right?

Tried to do a,b∈Z instead of N, cases a=0 (b is perfect square), and a=-1 (always works) are easy. For a ≤ b < 0, we can use similar approach as for N (obtaining only b=-1 solution), but I see no nice way how to solve case a ≤ -2, 1≤ b, because |4(a+b)| may be arbitrarily small (giving trivial solution a=-b, but not ruling out other solutions).

just switch to the point where you conclude that a has to be

Hello from Croatia. Which conference was it?

I've seen this implication in other videos and I don't see why. If n

Best shoutout ever seen 0:11

Maybe we could use the sophie germain identity?

Isn't a=3, b=6 a solution for the equation ab=2a+2b too?

I jumped in to the difference of squares and looked at factors of 4(a+b) and came to the same conclusions.

The first part is like a deus ex machina... more systematic approach would be rewrite n=(ab-k) with k>=1 (ab)^2-4(a+b)=(ab-k)^2 (ab)^2-4(a+b)=(ab)^2-2kab+k^2 ; cancel (ab)^2, rearrange 2kab=4a+4b+k^2 ; 2ab

Hay thanks for the Question LOVE FROM 🇮🇳DIA

Why can you say n

Nice video and problem, but if you're going to include 0 among the naturals, then (0,0) is a valid solution and you needed to take into account the case where a=0, which you didn't.

The question domain for n is confusing. Why couldn't it just say non-negative integers?

Excuse me, in 2:51 . . . what was the justification for jumping from n < ab to n ≤ ab - 1 ?

How did you guess that there must be only a few solutions to it?

20:07

3:50 I didn't get why we got the "ab-1" term

Let a+b=kl (i), ab=k+l (ii) where a,b,k,l are non negative integers. Then a^2b^2-4(a+b)=(k-l)^2. (i)-(ii) gives a+b-ab=kl-k-l => 2-(a-1)(b-1)=(k-1)(l-1)>=0 So, 2>= (a-1)(b-1)>=0 Suppose,(a-1)(b-1)=0 gives say a=1,then (k-1)(l-1)=2 => (k,l)=(3,2) or(2,3) and so (a,b)=(1,5) and similarly(a,b)=(5,1) By taking other two cases of (a-1)(b-1)=1,2 we get (a,b)=(2,3),(3,2),(2,2) solutions QED

I tried to do this question by myself. I did it this way: (ab)^2 - 4(a+b) = n^2 (ab)^2 - [2 * sqrt(a+b)]^2 = n^2 [ab - 2 * sqrt(a+b)] * [ab + 2 * sqrt(a+b)] = n^2 [ab - 2 * sqrt(a+b)] * [ab + 2 * sqrt(a+b)] = n * n u * v = n * n So I concluded that u = [ab - 2 * sqrt(a+b)] = n and v = [ab + 2 * sqrt(a+b)] = n So I concluded that u=v. (And that was bad assumption - I will write about this later below) u = v ab - 2 * sqrt(a+b) = ab + 2 * sqrt(a+b) -2 * sqrt(a+b) = 2 * sqrt(a+b) -4 * sqrt(a+b) = 0 sqrt(a+b) = 0 a+b = 0 b = -a In our original expression: (ab)^2 - 4(a+b) I substituted: b = -a and I get this: [a * (-a)]^2 - 4* [a + (-a)] [-a^2]^2 - 4* 0 a^4 - 0 a^4 And "a^4" of course is a perfect square since a^4 = (a^2)^2 But doing so I made a mistake. This is NOT so simple. I have made wrong assumption. I have overlooked such case when u is NOT equal to v but u*v IS a perfect squate, for example when u=4 and v=9, we get u*v = 4*9 = 36 = 6^2 so n=6 So we have the conclusion: Don't make the wrong assumptions. Don't overlook more complicated cases. Think: "Could there be some cases which I didn't think about?".

You're gonna have to post dog photos in the community tab now. ;p

Why the couple (0;0) is not included ?

In the last case of a=b=2, N=0, so not natural number?

if you assumed first n to be striclty positive, why not the second time?

Nice!

Im very bad at math. But it feels so cool to watch this😎

when ab = a + b + 1 ? ... if is it correct ... i have spent 5 minutes to think about it

for the case where n^2 + 8, and n^2 + 9, you could have noticed that the only square that is 8 less than another perfect square is n = 1, n^2 + 8 = 9, meaning (b - 2) = 3, b = 5; and for n^2 + 9 is a perfect square, that only works when n^2 = 16, or n^2 + 9 = 25, and then (2b - 1) = 5, so b = 3. In both cases it is easy to see that only one solution works, as the differences between consecutive squares are well known to be strictly increasing.

sir it can be solved by digital root method

Wait, 0 is not in N? Since when? Here in France we include 0 in N and usually note N* when we want only the positive integers. Is that a matter of country?

I love this problem , looks really exciting.

An excellent rationale to designate successor-closed integer subsets by placing the starting value as subscript to the N (e.g., N₀ for the "natural" numbers, N₁ for the "counting" numbers, N₂ for the "plural" numbers, etc. ad nauseam).

I’m from Ireland. Really cool to see this.

When a=-b?

4:44 please explain why he simply dropped the one half

The way he says Galway though

Wondering if there can be a geometric proof

Where is "Good place to stop"?

Can be solve easily by using discriminant of quadratic function.

You forgot a=0,b=0 if you include this it also works

2:49 Why is this true??

(a+b)^2-4ab=n^2 always sir. can we say that a+b =ab and it will always be true for given equation also😂😂😂 sorry just joking

Y not just a = -b? I didn't watch the video, but I was just intrigued by the thumbnail

When it's AJAR

9:37 oops does not work? this runs in milliseconds 10 for a=1 to 100:for b=1 to 100 20 dis=a*a*b*b:dis=dis-4*(a+b):if dis

if a =2, b=2, n = 0

Wow it was IMO

n cannot be zero as per the initial conditions

The dog is against considering 0 as an option for n

i am Chinese ，I also solved it, the method is a bit different

Really nice. But you forgot a=b=0. ;-)

I consider 0 in N, though...

Here's a fully deductive solution: First let's rule out the trivial solution (a,b;n) = (0,0;0). It can also be deduced but keeping a,b > 0 is helpful. (observe that if one of them is 0 and the other isn't then n² a+b (1) ab = a+b => a = b/(b-1) = 1+ 1/(b-1) => b=2 and a=2 this gives the second solution (2,2;0) (2) ab < a+b; by symmetry we can choose a >=b => ab < 2a so b k+m = a and km=a+1 => km = k+m+1, solving for k = (m+1)/(m-1) or k = 1+ 2/(m-1) => m = 2 and k=3 or m=3 and k=2 (an obvious symmetry) => a=5, b=1 this gives the third solution (5,1;1) (3) ab>a+b We don't need to investigate this case because by symmetry we can expect a=3 and b=2 (and k=5, m=1) (or you notice that in this case km < k+m and repeat the procedure in (2)) this gives the 4th solution (3,2;4) END PS: the first set of solutions can also be found by checking a=b => a^4 - 8a = n² => a(a-2)(a²+a+2)=n² if n=0 then a=0 or a=2 if n>0 => a|n and (a-2)|n => a²(a-2)²|n² => a(a-2)|(a²+a+2) => a|2 but this gives either 1 (no solution) or 2 (known solution)

26+6=1