There is a Surprising Answer to This Problem

[@richardfredlund3802]

that was fast. Really after seeing the solution it seems quite easy but I'm pretty sure if I tried it myself I wouldn't find it.

[@ΓιάννηςΚουρουμίδης]

I tried a different approach and ended up with at least 2 solutions (didn't prove that they were the only ones), (m,n)=(60,16) and (840,3136). However, when I put them back to the original equation I saw that the LHS was super close but not quite equal to 4.

[@Loots1]

So you didn't find a solution... if i saw m is 2 and n is 4 and say i found a solution am i rational?

[@Loots1]

If I say**

[@willbishop1355]

I used a quadratic equation about n instead of m: n^2 + (1-4m)n + m^2 = 0. After manipulating the discriminant, this implies that (4m+1)(4m-3) = (4m-1)^2, which is impossible.

[@MrLidless]

If negatives are allowed, (-4,-1) is the only solution.

[@bait6652]

guess teh q? is where in the author's proof would the neg soln have arised. His way of handling the radic square terms?

[@MrLidless]

@@bait6652 where n(3n-1) must be a perfect square. When n = -1, it is.

[@bait6652]

@@MrLidless yah where the author had a^2 * b^2 due to gcd, i prefer to use a^2*c and b^2*c

[@neladridinda4583]

A nice solution.. i have another approach with modulo: m²+n(n+1)=4mn Now taking modulo n we have m²=0(mod n) =>n|m clearly Now taking modulo m: n(n+1)=0(mod m) so n=0,1(mod m) now iff Case 1: n=0(mod m) Then m|n and n|m => m=n Putting this back in original eqn we get:- n=m= 0.5 (not an integer) Case 2:- n+1=0(mod m) So n|m and m|n+1 So m>n (we have already dealt m=n) And n+1>=m Only value satisfying these two inequalities simultaneously is n=m-1 But for that values, n|m => m-1|m which is impossible Hence, no solutions.. Please tell me if this is an erroneous proof..

[@Davs211]

"so n=0,-1(mod m)" incorrect. For example m=6, n=2

[@fransdebruijn99]

I tried it through number crunching using python on the PC which resulted in m= 7, n= 2, x= 3.9285714285714284 not quite 4 using integers for m and n. Below is a simple script m = 1 n = 1 for i in range(1,10): for j in range(1,10): x = (m/n)+((n+1)/m) if x3.9: print("m=",m,"n=",n," x=",x ) m+=1 n+=1 m = 1

[@euqed]

Please do something about the volume/sound quality of the videos, it would make them much clearer, thanks 🙏

[@CenturionDobrius]

I was about to say the same :)

[@capchemist]

I rather like the volume and sound quality of the videos

[@DaveyJonesLocka]

I wonder for which positive integer values is the expression on the left equal to an integer? There are some obvious low - valued solutions, but I wonder how many solutions there are?

[@sk8erJG95]

If m/n + (n+1)/m is an integer, then nm divides m^2 + n^2 + n, i.e. m^2 + n^2 + n = k(nm) for some integer k. Pulling the same trick as the video, write as a quadratic m^2 + (kn)m + (n^2 + n) and we see that (kn)^2 ‐ 4(n^2 + n) must be a perfect square. So (k^2 - 4)n^2 - 4n = n[(k^2 - 4)n - 4] must be a perfect square. As far as I can tell, this ONLY happens when k = 3. And the pairs (m,n) under 2000 that give this are (1,1), (2,1), (2,4), (10,4), (10,25), (65, 25), (65, 169), (442, 169), (442, 1156), (3026, 7921). Taking square roots of n gives the sequence 1, 2, 5, 13, 34, 89, which appears to be Fibonacci numbers! Unfortunately, I really can't figure out why n being every other fibonacci number means there exists an m so that m/n + (n+1)/m = 3.

[@sk8erJG95]

Those are all just calculations and data, but to be clear, it'd be nice to prove: 1) m/n + (n+1)/m = k only has solutions for k = 3. 2) When k = 3, all solutions (m,n) have n = F(2i - 1)^2 for positive integers i. 3) Explain why the heck this relates to fibonacci numbers at all

[@sk8erJG95]

Even on OEIS, I don't see anything about this on A081068 (n values) or A064170 (m values with m < n)

[@sk8erJG95]

Amazing! The reason is because the golden ratio phi satisfies the equation x^2 + 1/x^2 = 3. I'm not totally satisfied, but here is an asymptotic proof. Set m = 1 + F(k-1)^2 and n = F(k)^2, with k odd. Then m/n + (n+1)/m = (1+F(k-1)^2)/F(k)^2 + (F(k)^2+1)/(F(k-1)^2+1). As k goes to infinity, the first term goes to (F(k-1)/F(k))^2 ≈ (1/phi)^2. The second term goes to (F(k)/F(k-1))^2 ≈ phi^2, so the whole thing goes to 1/phi^2 + phi^2. But remember phi^2 = phi + 1 defines phi, so this is equal to 1/(phi+1) + phi + 1 = (1 + phi^2 + 2phi + 1)/(phi+1) = (3phi^2)/(phi^2) = 3. But the more wild thing is that this holds pointwise too. Very cool connection, prompted by your question! Thank you!

[@Bruno_Haible]

Here's a solution without discriminants. Put c = gcd(m,n) and write m=cp, n=cq with gcd(p,q)=1. The equation now reads cp² + q(cq+1) = 4 cpq. Hence c|q. Write q=cr. So, m=cp, n=c²r with gcd(p,r)=1. The equation now is p² + r(c²r+1) = 4 cpr. So, r|p². But since gcd(p,r)=1, r must be 1. So, the equation reduces to p² + c² + 1 = 4 cp. Since squares are ≡ 0 or 1 mod 4, the left-hand side is ≡ 1,2,3 mod 4, while the right-hand side is a multiple of 4.

[@EskilP]

Nice one

[@MohamedMahmoud-ey9tj]

I'm newbie correct me if my thinking is wrong, when i multiplied whole equation by m, it turned out everything is integer except m/n, implying m/n is integer, this says m is not smaller than n. Looking at original example therefore (n+1)/m is also an integer therefore n+1 is not less than m. The only way both those two statements hold is that n+1 = m and n=m , which is a contradiction unless n = 0 which it cannot be as the denominator

[@Yt-ff6hn]

Wow Indian question!

[@ignaciobenjamingarridoboba2071]

Well, i got it has not integer solutions too. Notice that (m^2+n^2+n)/n =4m, which means n divides m^2 and then n divides m (so m is multiple of n, m = q*n, with q belong to positive integers). Then you can rewrite the original equation as q+(n+1)/nq=4 and try to find values for n with pairs (q, (n+1)/qn) = {(1,3), (2,2), (3,3)}

[@antonioorlando5246]

Thanks for proposing this exercise. I don't know whether this solution method has already been suggested, but I proceed as follows: m/n + (n+1)/m is the sum of two positive numbers. Thus we need to analyze the following three cases: 1) m/n =1 and (n+1)/m=3; 2) m/n =2 and (n+1)/m=2; 3) m/n =3 and (n+1)/m=1. Now you can easily check that for each of these cases you don't find n,m\in\mathbb[N} and that's all folks!

[@mcwulf25]

Annoying. I guess we should always test for solutions before trying to solve. Can also do this mod 4. Perfect squares are 0 or 1 mod 4. If n is a p.s. then 3n-1 = -1 or 2 (mod 4).

[@alfreds1347]

1:23 Why does triangle is equal to square?:D

[@the-boy-who-lived]

By triangle it meant the discriminant. In quadraric equation there is a part like this √(b²-4ac) For the solution to the quadratic formula to be an integer b²-4ac (the discriminant) has to be a square.

[@mojota6938]

Oh, I didn't realize we were doing trick questions.

[@MizardXYT]

The closely related problem m/n + (n-1)/m = 4 has infinitely many solutions.

[@sk8erJG95]

Can you describe any pattern? Solution with m,n < 5000 are (4,1), (4,16), (60,16), (60,225), (840,225), (840,3136) And the also related problem m/n + (n+1)/m = 3 has infinitely many solutions, of the form (1+F(k-1)^2, F(k)^2) for F(k) fibonacci number and k an odd integer. There is also another solution for n = F(k)^2, which is m = (1+F(k+1)^2).

[@MizardXYT]

@@sk8erJG95 If you group the pairs by the n coordinate, you always get two points (m1, n) and (m2, n) (if you include the degenerate case of (m1, n) = (0, 1)). You could think of them as triplets, (m1, m2, n). All three follow the recursion f(k) = f(k-3) - 15f(k-2) + 15f(k-1), but with different initial conditions. Using them together, you can reduce the recursion depth. If you start with (m1 = 0, m2 = 4, n = 1), you can get the next triplet by (m1, m2, n) -> (4 n - m1, 15 m2 - 4 n + 4, 15 n - 4 m1 + 1) (0, 4, 1) (4, 60, 16) (60, 840, 225) (840, 11704, 3136) (11704, 163020, 43681) (163020, 2270580, 608400) (2270580, 31625104, 8473921) (31625104, 440480880, 118026496) (440480880, 6135107220, 1643897025) (6135107220, 85451020204, 22896531856) (85451020204, 1190179175640, 318907548961)

[@sk8erJG95]

@@MizardXYT Nice! How did you come upon that recursion? I'd like to try the same with the m/n + (n+1)/m = 3 problem to prove the Fibonacci solutions.

[@sk8erJG95]

@@MizardXYT Nevermind, I see! For m/n + (n+1)/m = 3, the triplets (m1,m2,n) pointwise satisfy a(n) = 8a(n-1) - 8a(n-2) + a(n-3), with initial conditions 1,2,10 for m and 1,4,25 for n. Then we get our infinite solutions via (m1,m2,n) --> (3n - m1, 8m2 - 3n - 3, 8n - 3m1 - 1). This gives the interesting identiies for Fibonacci numbers: 1 + F(k-1)^2 + F(k-2)^2 = 3F(k)^2 for odd k. And (1+F(k-1)^2)(1+F(k+1)^2) = F(k)^2(F(k)^2 + 1)

[@ariefa3544]

Why the fact that gcd equals 1 implies n & 3n-1 each is perfect square? I understand now, but you skipped that explanation.

[@dneary]

It would be interesting to explore when Pell type equations "m^2 - kn^2 = -1" have integer solutions for m,n,k integers. For m^2-kn^2 = 1 there are always solutions unless k is a perfect square, but apparently, with k=3, there is no solution for m^2-3n^2=-1 (a rearrangement of what you ended up with). But there are lots of values of k for which there are solutions - for k=5 we have 2^2-5(1^2)=-1 and for k=7, we have 8^2-7(3^2) = -1. What's special about 3 here?

[@dneary]

Oops... 8^2-7*3^2 is not equal to -1! It turns out that it is necessary for a solution that the continued fraction of sqrt(k) have an odd length (and I have no idea how you would prove that!)

[@factorial1059]

If i remember correcntly pell equation has infinitly solutions if and only if there exist atleast one solution

[@dneary]

@@factorial1059 Yes - and the negative Pell's equation has at least one solution when the continued fraction has an odd cycle length. Which is fascinating and very surprising!

[@factorial1059]

@@dneary can u explain in details what you mean by that?

[@dneary]

@@factorial1059 It's longer than a KZbin comment, but if there is a solution to a^2-kb^2=-1, then \sqrt{k} has a continued fraction of odd period. For example, \sqrt{13} = 3+1/(1+1/(1+1/(1+1/(1+1/(6+...))))) which has a period of 5, and the convergents of \sqrt{13} are 3,4,7/2,11/3,18/5,119/33,... and (18)^2-13(5^2)=-1.

[@factorial1059]

Or simply 0

[@Zachariah-Abueg]

why is it that any number that is congruent to 2 (mod 3) cannot be a perfect square?

[@Zachariah-Abueg]

i see now. if n ≡ 0 (mod 3), then n² ≡ 0 (mod 3). if n ≡ 1 (mod 3), then n² ≡ 1 (mod 3). if n ≡ 2 (mod 3), then n² ≡ 1 (mod 3). thus, for any natural number n, we have n² ≡ 0 or 1 (mod 3).

what about m = -4 and n = -1? Notice, if we substitute back, we have (-4)/(-1) + ((-1) + 1)/(-4) = 4 + 0 = 4

if it was all integers :)

[@timbenthall8681]

There is also a negative integer solution: n = -1, m = -4

[@jacobgoldman5780]

It says in question that m,n are both positive.

[@peamutbubber]

very nice method, can also be used in other nasty polynomial questions. Goodd videoo

[@SuperYoonHo]

Thanks

[@yoav613]

Well...i am not surprised

[@geoffreyparfitt7003]

Explaining why something is a surprise seems unusual but in this case that would be helpful.

[@sidimohamedbenelmalih7133]

3n²-n is a perfect square need to have a discriminant=0 Which is not the case

[@TheJSJosh]

But WolframAlpha says this problem has several solutions? For instance if m and n are both 0.5, 1 + 1.5/0.5 = 1 + 3 = 4

[@DaveyJonesLocka]

You are looking for positive integer solutions

[@dionisdsns1525]

But 4n could be =3n-1 and that be a square. N=-1 m=4

[@خلدوناللشوفي]

I send to you problem .would you solve it?

[@wernergamper6200]

no.

[@HAITRAN-cz5xm]

solution n=1/3, m=2/3

[@晓阳-d3p]

AMGM (n+1)/n at most 4 so 1/n at most 3 so 1/n =1,2,3 get n=1 only and so m+ 2/m =4 m^2 - 4m +2 = 0 no integer solution at all

[@nadkhaa8360]

i lv u 99.9%

[@joserego9681]

m=~ 7.16 n=2

[@peludogameplaysmachopeludo3506]

Integers

[@anantyadav4739]

X+1/x => 2 so Clearly this implies 1/m>2 which is not possible. One liner lol

[@DGramusset]

Check your inequations

[@MathsLycee209]

1/m

[@anantyadav4739]

What inequations?

[@getoutyoulol]

Yes even I did the same it's am>gm,u r right anant yadav.,,this was a one line question.

[@sk8erJG95]

If you replace the 4 with a 3, there are solutions. How does that affect your proof?

[@ready1fire1aim1]

Humanity has ten numbers (0, 1, 2, 3,...9) Newton: "0 is contingent" 🚫 and "1-9 are necessary" 🚫 (this is the basis of Newton Calculus/Physics/Geometry/Logic). Leibniz: "0 is necessary" ✅ and "1-9 are contingent" ✅ (this is the basis of Leibniz Calculus/Physics/Geometry/Logic). Is zero the most important number? Zero is the most important number in mathematics. Zero functions as a placeholder. Imagine a number, e.g., 5 and put as many zeroes behind it as you can think of. Zero drastically changes the value of the number from a mere 5 to 50, 500, 5000, 50000 and beyond. Which is the greatest whole number? There is no 'largest' whole number. Every whole number has an immediate predecessor, except 0. A decimal number or a fraction that falls between two whole numbers is not a whole number. Why is it impossible to divide by zero? The short answer is that 0 has no multiplicative inverse, and any attempt to define a real number as the multiplicative inverse of 0 would result in the contradiction 0 = 1. Is 0 a rational number? Yes, 0 is a rational number. Since we know, a rational number can be expressed as p/q, where p and q are integers and q is not equal to zero. Thus, we can express 0 as p/q, where p is equal to zero and q is an integer. Is 0 A whole number? The whole numbers are the numbers 0, 1, 2, 3, 4, and so on (the natural numbers and zero). Negative numbers are not considered "whole numbers." All natural numbers are whole numbers, but not all whole numbers are natural numbers since zero is a whole number but not a natural number. Why is 0 a good number? Zero helps us understand that we can use math to think about things that have no counterpart in a physical lived experience; imaginary numbers don't exist but are crucial to understanding electrical systems. Zero also helps us understand its antithesis, infinity, in all of its extreme weirdness. 🔘 ♾ ☯️ Do you agree with Newton that "0 is contingent" and "1-9 are necessary"? Then why are we learning Newton's Calculus/Physics/Logic? This is a problem 😕.

[@ayushparwal2004]

Am indian this is very easy question. I will say hint 3+1,4+0,0+4,2+2,1+3 only 5 possibilities.and then don't have common soln.means no solution

[@doodle1726]

You misunderstood the question bro

[@rassimTheBest3]

I found another way just notice that m²+n²+n=4mn Implies n|m² so n|m then m=d.n so we get s²n²+n²+n=4sn² then n²|n so n|1 implies n=1 now m²-4m+2=0 has no integers sol so we are done ✅

[@venkybabu8140]

When you cross multiply you get two squares. That means no solution.

[@خلدوناللشوفي]

I send to you problem .would you solve it?

[@خلدوناللشوفي]

The problem is the sum of floor function of the root (n)=[root1]+[root2].....+[root n] Find maximum value of n that make the sum prime number

[@خلدوناللشوفي]

The problem is the sum of floor function of the root (n)=[root1]+[root2].....+[root n] Find maximum value of n that make the sum prime number