A Diophantine Equation with Three Variables

This problem is from Moscow State University Entrance Exams
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[renyxadarox]

6:54 In general case the left fraction can be an improper fraction reducing to 30/13, so actually you need to solve the system of equations: xyz+x+z=30k, zy+1=13k, k∈ℤ, k≠0

[BarryMagrew]

The expression in x, y, z, on the left is a continued fraction. Therefore, a solution is only possible if the continued fraction expansion of 30/13 has exactly three terms. It does: 30/13 = 2+4/13 = 2 + 1/(13/4) = 2+1/(3+1/4). In standard continued fraction notation, 30/13 = [2; 3, 4].

[anastasissfyrides]

I didnt like this solution. It is unfinished. Firstly, you didnt take care of any negative solution possibility.Otherwise, why focus on the integers and not on the naturals? Secondly, how do we know there arent any other positive triplets as well?

[srijanbhowmick]

7:00 sorry to say but you cannot claim that numerator and denominator are equal to 30 and 13 respectively, they can also equal 60 and 26, 90 and 39

[MathElite]

Ooh interesting!
Congrats on 16K subscribers

[akshatjangra]

I think this one would be pretty easy by making cases

[ncantor]

The full solution is as follows:
First note from the original equation that no solution exists for y=0 or z=0.
Rearrange to get:
13x + 13z/(yz+1) = 30
Note that 13x and 30 are integers, so 13z/(yz+1) must also be an integer. And thus yz + 1 divides 13z
z and yz + 1 are coprime (by Euclidean algorithm), so yz + 1 must divide 13. Therefore yz + 1 = 1 or 13

yz + 1 = 1 implies that either y or z = 0, which we know cannot be the case so that case is ruled out. Thus yz + 1 = 13 -> yz = 12
z divides 12 and so ±z = 1, 2, 3, 4, 6, or 12
Plugging yz + 1 = 13 back into the top eqn: 13x + z = 30
Since -12 <= z <= 12, -12 <= 30 - 13x <= 12. Solving this inequality gives 18/13 <= x <= 42/13. Noting that x is an integer shows that x can only be 2 or 3.
Now we're ready to solve:
x = 2: 26 + z = 30 -> z = 4; 4y = 12 -> y = 3
x = 3: 39 + z = 30 -> z = -9 but -9 does not divide 12 so no sol'n.
Therefore x = 2, y = 3, z = 4 is the only solution

[travisgreen]

Continuing with the numerator and denominator equations, replacing 30 with 30k and 13 with 13k, substituting yz = 13k - 1 into x(yz + 1) + z = 30k yields x(13k) + z = 30k. Solving for z in term of k and z yields z = k((30 - 13x), Substituing for z in yz = 13k - 1 and solving for y yields y = (13k - 1) / [k(30 - 13x)]. For y to be an integer, k and (30-13x) must divide (13k-1). So (13k - 1)/k must be an integer. (13k - 1) / k = 13 - (1/k)., so for y to be an integer 1/k must be an integer; this only happens with k = 1 and k = -1. The k=1 case was illustrated in the video. Applying the same technqiue with factors of 14 when k=-1 has no solutions. So, (x, y, z) = (2, 3, 4) is still the only solution when the fraction is is 30k/13k where k != 0.

[satyapalsingh]

Very good method of teaching !!! Keep it up !!!

[warmpianist]

6:53 if that is not the case,
xyz+x+z = 30k, zy+1 = 13k for some integer k
From similar substitution, we get x*(13k)+z = 30k
This means z is divisible by k, let z = k*t for some integer t.
But from the second equation we will get 1 = k*(13-t*y), which only happens for k = -1 (already done k = 1) and t*y = 14
From z = -t and k = -1, substituting back to first equation we have
x*(-13)-t = -30
13x+t = 30 -> x = (30-t)/13
Since t*y = 14, we can try t = -14, -7, -2, -1, 1, 2, 7, 14 and none of them makes x an integer.
So there is only one solution when k = 1.

[udic]

You skipped a few checkings in both of the ways.
The solution is an integer.
Meaning it could also be negative.
In the first way 30/13 Is also equal 3 - 9/13 which might give another solution. It doesn't of course but you should have checked it.
The same goes for the second way.
When you check x=0, 1, 2 you should have checked also for x=3.
Again there are no more (integer) solutions but like you check for other positive values, you should have checked for the negative.
Otherwise it should have said that x, y, z in N+ (naturals plus 0 (because you checked x=0)) and not in Z (integers).

[pranavmeshram]

Rewriting the eqn in terms of y,
For y to be an integer, z|13x-30...(a) and 30-13x|13z...(b)
In (b), 30-13x|13....(i) or 30-13x|z...(ii)
For case (i), 30-13x=13, 1, -13, -1...this case gives no integer solution for x.
Thus case (ii) is valid...i.e., 30-13x|z and also z|13x-30(from (a))..this is only possible when z=30-13x
Putting z=30-13x in (*) we have,
yz=12.
Write 12 as a product of two integers and we get 12 cases, i.e, 12 (y, z) pairs.
Put each pair in original equation and check which pair gives x as an integer.
(Fairly quick to check all 12 cases)
Finally we get only one solution..
(x, y, z)=(2, 3, 4)

[GeorgeFoot]

For the second solution you should consider y and z both being negative. -12<=z<=12 and 13x+z=30 gives 2<=x<=3 so you ought to check x=2 and also x=3 to confirm it's not a solution.

[math]

great job Syber, thanks for solving this one

[pardeepgarg]

The easiest solution ever :
since y and z are integer greater than 1 so 1/(y+1/z) is less than 1
x + 1/(y+1/z) = 30/13 = 2 + 4/13
So x = 2
and
1/(y+1/z) = 4/13
(By comparing Integer and decimal part of both sides)
Now Taking reciprocal of 2nd
y + 1/z = 13/4 = 3 + 1/4
So y = 3 and 1/z = 1/4, z = 4
(By comparing Integer and decimal part of both sides)
The only solution :D









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[htp]

suggestion: dont premier so early it notifies your subscribers and they will likely leave early because it is going to take a lot of waiting, a time of like 15 mins early would be much more effective.

[sahilsinghbhandari]

This is easy question, it's ans will be 2, 3, 4. I solve it by splitting 30/13 in such a way that x will came integer and this gave me only one solution

[vuyyurisatyasrinivasarao]

Excellent easy but starting it's difficult...ur great sir...

[MathZoneKH]

Wow 3 variables in one equation !! You can solved, great

[aashsyed]

i like this channel so much you are growing so