I Solved A Quadratic Diophantine Equation | Integer Solutions?

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[luggepytt]

Here’s another approach.

a² + ab + b² looks a bit like (a+b)² = a² + 2ab + b²
Let’s try to utilize that somehow. Start by multiplying both sides of the equation by two:

2a² + 2ab + 2b² = 38
We can write that as a² + (a+b)² + b² = 38

So the problem boils down to expressing 38 as a sum of three squares.
38 = u² + v² + w²

We can easily find all possibilities if we observe that all the three numbers’ absolute values must be < 7.
Suppose |u| >= |v| >= |w|
Start with |u| = 6, and then work downwards:

|u|=6: 38 = 36 + 1 + 1 ⇒ {u, v, w} = {±6, ±1, ±1}
|u|=5: 38 = 25 + 9 + 4 ⇒ {u, v, w} = {±5, ±3, ±2}
|u|=4: 38 = 16 +16 + 6 ⇒ No solution, since 6 is not a square.
|u|=4: 38 = 16 + 9 + 13 ⇒ No solution, since |u| >= |v| >= |w| is not fulfilled.
|u|<=3: No solutions, since the sum will be <= 9 + 9 + 9 = 27

Now, {u, v, w} is really supposed to be {a, b, a+b}, meaning one of the numbers must be the sum of the other two. This disqualifies {±6, ±1, ±1}.
But from {±5, ±3, ±2} we can find a number of valid solutions.

(a, b, a+b) is any of:
(2, 3, 5),
(3, 2, 5),
(-2, -3, -5),
(-3, -2, -5),
(5, -3, 2),
(-3, 5, 2),
(-5, 3, -2),
(3, -5, -2),
(5, -2, 3),
(-2, 5, 3),
(-5, 2, -3),
(2, -5, -3)

Since the question was “what is a+b?”, the answer is:
±2 or ±3 or ±5

[antoniocarrillodiaz]

The question is to find a+b and is not necessary to calculate a and b previously
19 = a^2 + b^2 + ab = (a+b)^2 - 2ab +ab = (a+b)^2 - ab = (a+b)^2 - ((ab)^1/2)^2 = (a+b - (ab)^1/2)x(a+b + (ab)^1/2) Because 19 is a prime number we have
a+b - (ab)^1/2 = 1
a+b + (ab)^1/2 = 19
2(a+b) + 0 = 20 then a+b = 10 The other answer is obtained with -1 and -19, resulting a+b = -10

[thichhochoi]

76-3b^2 must be a perfect square and b <6. We have b = 5, 3, 2. No need to use MOD 3

[chiahungli]

The last equation should be a^2+5a+6=0 instead.

[fadetoblah]

Since a and b are technically interchangeable in the original equation, if you haven't found values of ±1 or ±6 for b, then you should certainly not end up with values of ±1 or ±6 for a. Also, and for similar reasons, you should not disregard negative values of b, since the set of possible b values must mirror the set of possible a values.

[benjaminvatovez]

Thank you for this video. I used these 2 equivalent equations:
(a-b)²+3ab=19 for ab>=0 (let us suppose a>=b>=0 without loss of generality)
(a+b)²-ab=19 for ab<=0 (let us suppose a>=0>=b without loss of generality)
Now we can consider both cases one by one quite fast as the square number and the other term are both non-negative:
3ab=19-0² (not a solution as ab must be an integer)
3ab=19-1²=18 then ab=6 though a+b=1 thus (a, b)=(1, 0) then ab=0 contradiction
3ab=19-2²=15 then ab=5 though a+b=2 thus (a, b)=(2, 0) or (1, 1) then ab is not 5: contradiction
3ab=19-3²=10 (not a solution)
3ab=19-4²=3 then ab=1=1.1 and a+b=4: contradiction.
3ab-19-5²<0 no solution from here.
For |ab|=19-(a+b)² = 19, 18, 15, 10, 3, this is the same process.

[echandler]

A different way to look at the problem. Also different from the historical methods used to solve these problems. We only need to find non-negative solutions where a <= b. The rest can be generated by swapping a and b, negating a and b or taking a+b, -a. (or some combination of these). Each (a, b) found gives rise to up to 11 other solutions.

[pageegap]

Starting from a²+ab+b²=c (where c=19) we are searching for sums (a+b), a & b are integers.

we have (a+b)²-ab=c
and 4ab = (a+b)²-(a-b)²,  
then (a+b)²-¼(a+b)²-¼(a-b)²=c
then ¼(a-b)²=c-¾(a+b)² 

so (a-b)²=4c-3(a+b)².

As (a-b)² is an integer squared we have the property P(n)=4c-3(n)² is also an integer squared for n=(a+b)

Then we have to find n verifying this property for c=19 which happens when |n|<=5 because 3×(6)²=108 is greater than 4c=76.

We find that P(5)=1, P(3)=49=7² or P(2)=64=8².

So finally the set of solutions for (a+b) is S(a+b)={-5; -3; -2; 2; 3; 5}

Note: if we want to find also the values for a and b we can investigate for b satisfying the second degree equation in "a" then find the a-roots. We can observe that the discrimant of such eq. is given by ∆=b²-4x1x(b³-c)=4c-3b² and this discriminant must be an integer squared to obtain integer roots. As we can see ∆ is verifying the same property as above for b : P(b)=4c-3(b)². So the set of solution S(b) for b is the same set of solutions S(a+b) for a+b 

so S(b)=S(a+b).

[swthiel]

I believe Z for integers comes from German "Zahlen" = numbers

[bsmith]

Let's start by completing the square.
a^2+ab+b^2=19
4a^2+4ab+4b^2=76
(4a^2+4ab+b^2)+3b^2=76
(2a+b)^2+3b^2=76

3b^2<=76, which then means |b|<=sqrt(76/3)=5.033. So then with b needing to be an integer we must have b from the set {0, +/-1, +/-2, +/-3, +/-4, +/-5}. Over these values 76-3b^2 is a perfect square for b=+/-5, b=+/-3, and b=+/-2
b=+/-5 -> (2a+b)^2=1 -> 2a+b=+/-1. Then (a, b) can be (2, -5), (3, -5), (-2, 5), or (-3, 5). Then a+b=-3, -2, 2, or 3.
b=+/-3 -> (2a+b)^2=49 -> 2a+b=+/-7. Then (a, b) can be (-2, -3), (5, -3), (-5, 3), (2, 3). Then a+b=-5, -2, 2, or 5.
b=+/-2 -> (2a+b)^2=64 -> 2a+b=+/-8. Then (a, b) can be (5, -2), (-3, -2), (3, 2), (-5, 2). Then a+b=-3, -2, 2, or 3.

Then collecting all the possible for a+b we have a final solution set of a+b is in the set {-5, -3, -2, 2, 3, 5}.

[cameronspalding]

@ 5:55 The reason why Z is the symbol for integers is because of the German word Zalen which means number, back in the day the word number was meant to mean integer.

[someperson]

(2a+b)^2 +3b^2 = 4(a^2+ab+b^2) = 4(19) = 76. Hence, b^2 <= 76/3 < 26. Thus, |b| <= 5. This gives 11 possible values for b. Check these, we find the only solutions are:
(a, b) = (2, 3), (-2, -3), (3, 2), (-3, -2), (2, -5), (-2, 5), (5, -2), (-5, 2), (3, -5), (-3, 5), (5, -3), (-5, 3).
{a + b| a^2 + ab + b^2 =19, where a, b in Z} = {-5, --3, -2, 2, 3, 5}.

[faridmohammadmaalekghaini]

If the equation is multiplied by a-b, then we obtain:
a^3-b^3= 19(a-b),
and if we want natural pairs of solutions it will be evident that (a, b)=(3, 2) is the unique solution.

[wryanihad]

Take this example and make a vedio on it. Because no program can solve it !!!

The question is xy=x

Simple but a little hard
I hope you will accept my opinion ❤❤❤

[Mekenyejustus]

Clearly, a=2 when b=3 and a=3 when b=2 therefore, a+b=2+3=5

[dariosilva]

(5, -6) and (5, 1) are incorrect solutions. But (5, -2) and (5, -3) should be on the list instead.

[mbmillermo]

I think it would have been quicker to just plug b values into the discriminant because they had to be no bigger than 5. You could have skipped the whole b, c thing.

[anandmehta]

The original equation is symmetrical in a and b so the solutions MUST also be symmetrical.

[cameronspalding]

@ 4:12 why do you write minus-plus rather than plus-minus

[adityasingh]

Amazing, but please solve that a + b. 😂😂