Diophantine Equations! Can You Solve this System? | Simple & In-Depth Explanation

[jimmykitty]

*Wow!!! What a fantastic mathematical problem it is!! Elegantly explained Boss* 😎 *Love from Bangladesh* 🇧🇩❤

[pierreneau]

Other option, this system can be solved classically by substitution: Eq1 gives x = (11+y)/(y-1) and Eq 2 gives z = (14+y)/(y-1). So you replace x and z in Eq 3 and you obtained a second degree equation in y : y2 - 2y-8 = 0 and y=-2 (to be eliminated if y is positive integer) and y = 4 and so x = 5 and z = 6. Thank you to share nice math exercise.

[wangwenho]

Taiwan student here! I have more efficient solution. SOL: ab*bc*ac = 3600 = (a*b*c)^2 --> a*b*c = 60 and we can use it divide by ab, bc, ac respectively to get c, a, b respectively.

[dreael]

The tree equations 7 to 9 can be solved easily without a trial and error table: #7 can be converted into a=12/b (#7.1). In #9, I replace a so (12/b)*c=20 result => *b both sides => 12*c =20*b => b=(12/20)*c = (3/5)*c (#9.1) => Substitution of b in equation #8: (3/5)*c*c=15 => c^2=25 => c=+/-sqrt(25)=+/-5 => substitution in #9.1: b=(3/5)*(+/-5)=+/-3 => Substitution in #7.1: a=12/(+/-3)=+/-4 There are two result sets actually: { a=4, b=3, c=5 } and { a=-4, b=-3, c=-5 } because we needed sqarerooting one time only. Because of x=a+1, y=b+1 and z=c+1, we then get both the result sets: { x=5, y=5, z=6 } and { x=-3, y=-2, z=-4 } To get the full right result, the second result set (mathematically also correct) must be rejected because we want those solutions with positive numbers only, so { x=5, y=5, z=6 } remains as final result.

[tmacchant]

Multiply equations ab=12, bc=15, ac=20, then (abc)²=3600, thus abc=±60. a=abc/bc=±60/15=±4, b=abc/ac=±60/20=±3, c=abc/ab=±60/12=±5.

[mikezilberbrand]

ab*bc*ca = (abc)^2= 12*15*20, so abc=60, then divide 60 by 12, 15, 20 to get c, a, b.

[Gargaroolala]

Omg I can’t believe I solved this qns via the same method as u! I got equations 4, 5, 6 (same as yours). I didn’t do guess and check. I just solve them via substitution to get the final answer. 🙂

[quinntalley]

I enjoy your videos very much and this one is no exception. You take a rigorous approach to solving this problem but I wanted to point out a far easier method. Consider for each of these equations an almost trivial question: What pairs of positive integers satisfy the condition that their product minus their sum equals 11, 14 and 19? [Since these equations may be rewritten as the product minus their sum: xy-(x+y)=11, yz-(y+z)=14 and zx-(z+x)=19] Clearly the pair (4, 5) gives the result 4*5-(4+5)=11. The pair (4, 6) gives a result of 4*6-(4+6)=14. And the pair (5, 6) gives a result of 5*6-(5+6)=19. Now, associating the common integer in first two pairs with y, y=4. Likewise, associating the common integer in the first and third pairs with x, x=5. And, finally, associating the common integer in the last two pairs with z, z =6. Basically, you can solve this almost by inspection. Thanks again for all your great videos.

[bentels]

Very enjoyable problem, nice puzzle!

However, you can be a bit smarter about equations 7, 8 and 9 and save yourself some work:

ab = 12 AND ac = 20
Therefore a divides 12 and 20.
Their common divisors are 1, 2 and 4, with the GCD being 4.

Pick 4 for a, you find b and c. The value 3 for b is also GCD(12, 15). And the value 5 for c is also GCD(15, 20). So this solution is unique for a, b, c IN N.

[math]

watching and learning, thanks for sharing this system of diophantine equations

[johnbrennan]

Z=19+x\x-1. Z=14+y\y-1 and y= 11+x\x-1. That gives 19+x\x-1 =14+y\y-1. Substituting 11+x\x-1 for y and using some algebra I ended up with 15x^2-30x-225=0.So x= 5 and-3. But x is a positive integer so x=5 etc.

[ManjulaMathew-wbzn]

I figured you didn’t have to assume a B and c are integers.
If you multiply equations 7 and 8 and divide by 8. You get b^2=9. If you want a B and c positive you get B=3. Equations 7 and 8 gives a to be 4 and c to be 5.
Also note the negative of those values also satisfy the original equations.
The fact of the matter is even if the RHS ended up being different values they are still solvable but the answers may end up being fractions.
I really liked your approach on the rest of the solution.

[robertthomson]

Could of used the fact that b had to be a an odd number due to equation 8 of bc being odd therefore both b and c are odd. Would of ment not checking as many possibilities on with equation 7

[nicogehren]

great approach sir and very well explained thanks and keep rocking!

[zakinaqvi]

Multiply equations 7 and 8. The result will be ab^2c=180. Now put value of ac from equation 9. The value of b will be +/-3. The values of a and c will also be calculated in the same manner. Thanks.

[diamonddave]

The negative solutions are x = -3; y = -2 & z = -4

[TPinesGold]

Simpler from equations 7, 8, 9 ...
A = 12/B
C = 15/B
AC = 180/B² = 20
B² = 9
B = 3
A = 4
C = 5
etc

[EngMorvan]

@6:06, it would be faster if u pick eqn 8 instead of 7, bc 15 has less prime factors than 12 or 20 so the list of possible products is smaller.

[johnbrennan]

Thank you for such a great solution--- very interesting indeed.

[buffuniballer]

Once you get to the series of equations of ab=12 bc=15 and ac=20, it's pretty easy to see that c is going to be 5. So it's worth taking a chance that c=5 and then work out a and b to see if they fit.

Turns out they do.

From there, work out the values of x, y and z and Bob's your uncle.