find the integer solutions of this cubic 

I've been following Prof Penn for a fairly long time, and I can't remember the last time he did polynomial long division... takes me back many decades

Here's my solution: Start by factoring as (x-y)(x^2+xy+y^2)=xy+61, x^2+xy+y^2 can be written as 1/2(x+y)^2+1/2(x^2+y^2) which is clearly greater than 0 Case 1: x-y=0, this clearly has no solutions as it reduces to x^2+61=0 Case 2: x-y>0, since x and y are integers, x-y is greater than or equal to 1 so x^2+xy+y^2≤xy+61, cancelling gives x^2+y^2≤61, checking the finite number of solutions gives (6,5),(-5,-6) as the only ones Case 3: x-y

An intuitive substitution is u=x-y, v=xy. This leads to 3u-1 having to divide 1646=2*823 (semiprime). This creates only 4 cases for u, just 2 of which are integers. One case gives both solutions because the substitution is quadratic, the other fails produce a rational x.

Another (fairly unintuitive) solution, is to multiply both sides by 27 and subtract 1 from both sides and you get (3x)³ + (−3y)³ + (−1)³ − 3(3x)(−3y)(−1) = 1646. The LHS is in the form a³ + b³ + c³ -3abc which famously factorises as (a + b + c)(a² + b² + c² - ab - bc - ca) and so the LHS can be factored as such. Then it's just matching factors that work.

14:53

Nice and interesting video. Could you please make more videos about number theory or Diophantine equations?

in order for dividing by 3b-1 to be allowed it must not be zero or 3b=1 must not hold. since b must be an integer 3b=1 cannot hold. so dividing is fine. I felt it would be important to point that out to justify the division

Hi, Michael your method is amazing and beneficial. Here's my method which is easy method:set x=y+m it's giving a polynomial p(y)=(y^2)(3m-1)+3m(m-1).y+m^3-61=0 let calculating Delta we find Something in terms of m=a perfect square ⬛️ >0 so then if m>6 delta is

You can represent 61 as (4)^3-3 and find that for positive arguments the equation is solved for y with a remainder of 2 when divided by 3 and x divisible by 3, and immediately find (x;y) equal to (6;5), prove that for large x,y the equation has no solutions, because the left part will invariably be larger than the right. Cases with negative arguments are easy

why the maximum power is 3 not 2 .There is no general solution to this type of problem

If you look at the factorization of the difference of cubes you can make some inferences. (x-y)(x^2 +xy +y^2) = xy + 61 If you assume (x-y) is an integer since it must be due to the nature of the problem, you can do the following cases Case 1: (x-y)=1 (x^2 +xy +y^2) = xy + 61 x^2 + (x+1)^2 = 61 Which or knowing it's a sum of squares what gets you 61? 5^2=25 and 6^2=36. Also realizing that squaring negative numbers also works, but the difference condition requires x to be greater than y. Case 2: (x-y)=2 2(x^2 +xy +y^2) = xy + 61 2x^2 + 2y^2 = 61 - xy x^2 + y^2 = 61/2 - xy/2 Where we see that x and y can't be 0 because you can't have a sum of squares that gives you 30.5. Then we know the minimum it can be is X=3, y=1, which gives us 10 ≠ 29.5 Incrementing X=4, y=2, yielding 20 ≠ 26.5 X=5, y=3, but this goes over at 34 ≠ 23 So there is no solution at (x-y)=2 We can keep trying the cases but we end up with (x-y)=z x^2 + y^2 = 61/z +xy* (1-z)/z Where we know that x or y can't be 0, z can't be zero. We also know that there is an answer for z=1 where x=6 so the integers can only go from 0 to 6 or their negative counterparts. It'd be interesting to try where Z is less than 0.

a similar proof but much less elaborate is to replace x with y +m where m is an integer and then proceed to show that permissible values of (3*m-1) are the factors of 1646 which are 1,2,823 and itself. m can only be equal to 1 and then one can easily arrive at your final solution

I often go brain cramping when I see a long division done this way, because I’ve learned it the French way with bracket inverted and the divisor in the bracket.

This can be factored, motivated by those two cubes into the form a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca). After multipying by 27, and using the formula for a=3x, b=-3y and c=-1, this factors as (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca) = 27*61 - 1. Now, this again doesn't seem to have any nice way of solving, but the brute force is not so bad. Firstly the "bigger" parentheses is always positive and therefore the other one has to be positive. Also the first one is smaller than the second one so it leaves us only a few cases, again a lot of work, but the problem doesn't have any "nice" solution.

I did this by putting y=x+k, then few things cancel out and we are left with quadratic, where y is variable and k is parameter. We calculate delta and notice only for k=1,2,3,4,5 it is positive (some boring analysis- we factor out (3k-1)*(cubic trash) and then look at derivative) but even more- straight checking says only for k=1 it is perfect square. Therefore using quadratic formula we get y=5 or y= (-6) which leads us to final solution

In what's almost the same, can choose integer a s.t. x = y + a in order to kill the y^3 term, leaving a quadratic in y with coefficients in a. Even though a is unknown, can in principle solve that quadratic for y, and a & y both being integers creates hard constraints on their possibilities. Given x,y integers s.t. x^3 - y^3 = xy + 61, let a = x - y. Then 3a y^2 + 3a^2 y + a^3 = y^2 + ay + 61, so (3a - 1) y^2 + (3a^2 - a) y + (a^3 - 61) = 0. Thus y^2 + a y + Q = 0, where Q = (a^3 - 61)/(3a - 1). a & y integers => Q an integer. Q = (a^3 - 61)/(3a - 1) = a^2/3 + a/9 + 1/27 - (1646/27)/(3a - 1), so 27 Q = 9a^2 + 3a + 1 - 1646/(3a - 1). Since everything there is an integer, and (3a - 1) | 1646, so using 1646 = 2(823), 823 a prime, gives: (3a - 1) in { 1, 2, 823, 1646 }. Thus (3a - 1) in { 2, 1646 } via congruence mod 3, so a = 1 or a = 1647/3 = 549, so a in {1, 549}. Have y^2 + a y + Q = 0 where Q = (a^3 - 61)/(3a - 1), and a in {1, 549}. If a = 549, then 3a - 1 = 1646, so Q = ((549)^3 - 61)/(3(549) - 1) = 100,528. Then y^2 + 549 y + 100,528 = 0. Discriminant = 549^2 - 4(100,528) = 10^4 { 5.49^2 - 4(10.0528) } < 10^4 { 6^2 - 4(10.0528) } < 0, so 2 complex roots, so a = 549 can't work. If a = 1, then Q = ((1)^3 - 61) / (3(1) - 1) = -60/2 = -30, so y^2 + y - 30 = 0, so (y - 5)(y + 6) = 0. Thus y in { 5, -6 }, and since x = y + a, get two possible solutions y=5, x=6 or y=-6, x=-5. Checking: x^3 - y^3 ?=? xy + 61 6^3 - 5^3 ?=? (6)(5) + 61 216 - 125 ?=? 30 + 61 216 - 125 = 91. x^3 - y^3 ?=? xy + 61 (-5)^3 - (-6)^3 ?=? (-6)(-5) + 61 6^3 - 5^3 = (6)(5) + 61 (as before). SOLUTION: y=5, x=6 or y=-6, x=-5.

Easily xy0. If (x,y) is a solution then clearly (-y,-x) is too. But if xy0. But in this case, clearly x>y. Since x2-xy+y2>xy, we have xy+61 > (x-y)xy. We can consider 2 cases: 1) x-y>1 => xyy y2+y-30=0 and hence the only solutions are (6,5) and (-5,-6).

Wake up early morning and watch a math video. Keeps you alert while day❤

I think these number theory problems start to get boring. Maybe it's better to focus on more advanced mathematics, not only highschool problems

My first thought was "complete the cube" and hope to get some kind of factorisation on the left and a number on the right. Completing the cube gives: (x - y)^3 = -3x^2.y + 3x.y^2 + xy +61 = 61 - xy.[3(x - y) - 1] Subtract 1 on the left and it factorises as [(x - y) - 1][ some other junk]. Hmm, that (x - y) - 1 is kinda like 3(x - y) - 1 on the right. 💡Multiply by 27 before subtracting 1, gives (after a little rearrangement): [3(x - y) - 1][9(x - y)^2 + 3(x - y) + 1] + 27.xy.[3(x - y) - 1] = 27.61 - 1 = 1646 Or: [3(x - y) - 1][9(x - y)^2 + 3(x - y) + 1 + 27.xy] = 1646 Which is the factorisation we need Thus 3(x - y) - 1 = (+/-)1, (+/-)2, (+/-)823, or (+/-)1646. Solve for x - y (which must be an integer), then find x (and hence y) from 9(x - y)^2 + 3(x - y) + 1 + 27.xy = quotient of the value of x - y in 1646 Sorta like bo luo's solution below, only a lot more pedestrian

multiplying by 27 was a bit like magic

Multiplying by 27 out of nowhere ^^ pretty creative

Thank you Micheal !!

From (x-y)(x^2+y^2) = 61 => x^2+y^2 x^2 x = -5 => (x^2+6) = 31 is a solution

Well done.

Not related to this but are you going to cover the new proof of the Pythagorean theorem that is supposed to be non-circular?

x^3 - y^3 =xy+61 implies (3x-3y-1)(+ve factor in x, y) = 2x823, 3x-3y-1=2 gives the two solutions. No integer solution for the other factor.

Why must 2 mod 3 apply?

Isn't b a integer ? So u have to consider the negative factors of 6584 .

One thing you should ALWAYS do at the end of these videos is double check all solutions against the original equation.

Nice solution :)

Another way is possible to solve the equation y^3-x^3 = y*x-61 =0 in integers. It is not difficult to see that the variables x and y must have different parity: if x is even, then y is odd, and vice versa. Then, if we write y=x+a, then a must be an odd integer. Substituting this into the original equation, we get a quadratic equation with respect to x with the parameter a: (3*a+1)*x^2 +(3*a^2 +a)*x+a^3 +61 =0. Its discriminant D(a)= -3*a^4 +2*a^3 +a^2 is 732*a - 244. Examining D(a), for example, graphically, we find that D(a)>0 for odd integer values a = -1, -3, -5, but only for a= -1 is the square of an integer. So, a= -1. The quadratic equation is reduced to the form x^2 - x-30=0. x=6, y=5 & x= -5, y= -6.

Thanks for the video i was stuck in a similar problem now i know how to tackle

Thank you, professor.

First! Great solution, too.

ah yes, the solutions of a function

@M. Penn: If you're multiplying both sides by 4, shouldn't the initial b on the lefthand side have a 4 in front of it, or am I mistaken?

Why can’t 3b - 1 be a negative integer?

What if 3b - 1 is negative?

Hi Dr. Penn!

why dont chack negative options for 3b-1? Like -1,-2 ...

Just for fun I asked GPT4 to solve the problem, and it came up with the same answers in a far easier way, but I think it made an invalid assumption. It did the first step the same -> (x - y)(x^2 + xy + y^2) = xy + 61 But then it did two cases: Case 1 : x - y = 1 -> solving for x and y gave (x, y) = (6, 5) and (-5, -6) Case 2 : x^2 + xy + y^2 = 1 -> solving for x and y gave (x,y) = {(1,0),(0,1),(-1,0),(0,-1)} It then tested both cases and found case 2 offered no valid solutions. I think it assumed that xy+61 was prime somehow.