$a^3+b^3+3\big(a^2+b^2\big)=700(a+b)^2$

Let $a$ and $b$ be rational numbers such that $a \ge b,$ their sum is an integer, and the equation above is fulfilled. Find all solutions $(a, b)$ and enter your answer as $\sum (a+b)$ .

The answer is 5586.

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Nice approach using the discriminant! I arrived at the same equation thinking "difference" rather than "discriminant":

Write $s = a + b$ and $d = a - b$ and translate the problem into $(\tfrac14s^3 + \tfrac34sd^2) + (\tfrac32s^2 + \tfrac32 d^2) = 700s^2,$ which then becomes $s^2(2794 - s) = 3d^2(s+2).$ This approach avoids calculation of the product $ab$ , which is not really needed here and gets transformed into $a - b$ when calculating the discriminant.

The problem is now indeed finding under what conditions the ratio of
$2794 - s$
and
$3(s+2)$
is a square. I like the nice trick of
*
multiplying
*
numerator and denominator: if
$D = ax^2$
and
$N = ay^2$
then
$DN = a^2x^2y^2$
is a square. I had not thought of this, and was therefore stuck with considering the possible values of the gcd of the two terms.

Sadly I missed the obvious case $s = 2794$ in submitting my answer :(

Arjen Vreugdenhil
- 3 years, 12 months ago

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You and Kazem both considered $a-b$ , and we all obtained the same equations. Calvin's comment is that $a-b$ comes "out of thin air", while the discriminant approach gives a pretty algorithmic route through problems of this type.

It is nearly always easier to consider integer Diophantine equations than rational ones!

Mark Hennings
- 3 years, 12 months ago

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A disagree with $a - b$ coming "out of thin air".

In a way I simply applied a linear change of coordinate basis: $(a, b) \mapsto (a + b, a - b)$ . It is a standard approach in multivariate problem solving to choose such a transformation.

Another way of looking at it: the sum $a + b$ and the difference $a - b$ are the simplest expressions that are symmetric vs. anti-symmetric under swapping $a$ and $b$ . This problem is obviously symmetric in $a, b$ , except for the anti-symmetric requirement $a > b$ .

To me, this choice is also intuitive. When I see a problem symmetric in $a$ and $b$ , I immediately wonder about the average (half-sum) and half-difference. A very simple example is the mental calculation of $17\times 23 = 20^2 - 3^2$ , where $a = 17$ and $b = 23$ have average $20$ and half-difference $3$ .

Even when solving $x^2 + Bx + C = 0$ , $x = -\tfrac 12 B \pm \tfrac 12\sqrt{B^2 - 4C} = \tfrac12(a + b) \pm \tfrac12(a - b).$ Your approach of finding $ab = D$ then using the determinant formula, boils down to calculating the squared difference $D = (a - b)^2$ .

Moreover, knowledge of sum and difference makes it very easy to recover $a = (s + d)/2$ and $b = (s - d)/2$ , which is more difficult when one knows only the sum and the product.

Arjen Vreugdenhil
- 3 years, 12 months ago

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@Arjen Vreugdenhil – To clarify, it "comes out of thin air" in the sense of "Why is this the exact term that we want to consider"? Why didn't we look at say $a + 2b$ or $2a - 3b$ ?

I have nothing against "I was hit by inspiration to consider this term". However, when there is a motivating reason behind it, I like to highlight it.

Sometimes, inspiration comes from experience, and we might find it hard to verbalize. E.g. mentioning that "For symmetric equations, it is often helpful to consider the change of variables $(u,v) = ( a + b , a - b )$ ", would help to reduce the mystique behind the solution.

Your approach for the first part of the problem is nicest.

Kazem Sepehrinia
- 3 years, 11 months ago

Check your addition at the end. 2794 + 697 + 2095 = 5586, not 5588.

Stacey Greenstein
- 3 years, 12 months ago

Isn't 2792n instead an -2792n at the equation about n and k?

세연 조
- 3 years, 11 months ago

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Sorry i meant -2792n instead of 2792n

세연 조
- 3 years, 11 months ago

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Whoops the first comment was the right thought of mine... really sorry about that.

세연 조
- 3 years, 11 months ago

Can u please clear that last part of quadratic residue?

Arijit Dey
- 3 years, 10 months ago

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Check out the Brilliant wikis on quadratic residues and the Legendre symbol .

Mark Hennings
- 3 years, 10 months ago

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I did , but the in the first line where you used (-3|233) ; where did this (-3) come from???

Arijit Dey
- 3 years, 10 months ago

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@Arijit Dey – Note that $u^2 + 3v^2 \equiv 0 \pmod{233}$ , and $233$ is prime. If $233$ does not divide $v$ , then we can find an integer $w$ such that $vw \equiv 1 \pmod{233}$ , and hence $(uw)^2 + 3 \equiv 0 \pmod{233}$ , which implies that $-3$ is a quadratic residue modulo $233$ . Since this is not true, it follows that $233$ divides $v$ , and so on...

Mark Hennings
- 3 years, 10 months ago

Let $k=a+b$ and rewrite the equation: $k^3-3abk+3(k^2-2ab)=700k^2 \ \ \Longrightarrow \ \ ab=k^2 \frac{k-697}{3k+6} \ \ \Longrightarrow \ \ a^2-ka+k^2 \frac{k-697}{3k+6}=0 \qquad (1)$ This is a quadratic in $a$ and hence its discriminant must be a square of a rational. Thus $\Delta_a=k^2-4k^2 \frac{k-697}{3k+6}=k^2 \frac{2794-k}{3(k+2)} \qquad (2)$ is square of a rational and consequently $\frac{2794-k}{3(k+2)}$ is non-negative and square of a rational. Observe that $-1 \le k \le 2794$ . Check that for $k=-1$ the expression $\frac{2794-k}{3(k+2)}$ is not square of a rational. For $k=0$ and $k=2794$ , discriminant becomes zero and solving quadratic equation $(1)$ together with $k=a+b$ yields to two solutions from here: $\color{#D61F06} (a, b)=(0, 0), (1397, 1397)$ .

Henceforth, $0<k<2794$ and for co-prime positive integers $p, q$ $\frac{2794-k}{3(k+2)}= \frac{p^2}{q^2} \ \ \Longrightarrow \ \ k=\frac{2794 q^2-6p^2}{3p^2+q^2}$ Thus, $3p^2+q^2 \mid 2794 q^2-6p^2$ or $3p^2+q^2 \mid 2794 q^2-6p^2+2(3p^2+q^2)=2796 q^2 \qquad (3)$ Notice that $d=\gcd(q^2, 3p^2+q^2)=\gcd(q^2, -q^2+3p^2+q^2)=\gcd(q^2, 3p^2)=\gcd(q^2, 3)=1, 3$ And $d=3$ occurs when $3 \mid q$ . Suppose that $d=1$ . Then from relation $(3)$ we have $3p^2+q^2 \mid 2796=3 \times 4 \times 233$ , but $3 \nmid q$ , and this implies that $3p^2+q^2 \mid 932=4 \times 233$ . Hence possible values for $3p^2+q^2$ are $4, 233, 466$ and $932$ . We claim that $3p^2+q^2=4$ . We use the facts that a perfect square can only be congruent to $0, 1$ mod $4$ and congruent to $0, 1, 4, 7$ mod $9$ . Note that $3p^2+q^2$ mod $4$ never produce $2$ and, thus, $3p^2+q^2=466$ is not possible because $466 \stackrel{4}{\equiv} 2$ . Also, $3p^2+q^2$ mod $9$ never produce $5$ and $8$ . Hence $3p^2+q^2=233$ and $3p^2+q^2=932$ are not possible because $233 \stackrel{9}{\equiv} 8$ and $932 \stackrel{9}{\equiv} 5$ . So $3p^2+q^2=4$ and $p=q=1$ , which gives $k=697$ and equation $(1)$ gives another solution, namely $\color{#D61F06} (a, b)=(697, 0)$ .

Remaining case is $d=3$ , where $q=3q'$ for a positive integer $q'$ . Then relation $(3)$ gives $3p^2+9q'^2 \mid 2796 \times 9q^2.$ And since $3 \nmid p$ , we get $3p^2+q'^2 \mid 4 \times 233$ . By the same arguments above we get $p=q'=1$ . Therefore $p=1$ and $q=3$ .This yields to $k=2095$ . Using equation $(1)$ gives the last solution $\color{#D61F06} (a, b)=\left(4190/3, 2095/3 \right)$ .

Then the final answer to this problem is $\sum k= 0+2794+697+2095= \boxed{5586}$ .

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in the equation 3 I think it will be 2796q^2

If $3p^2+q^2=1$ and $p=q=1$ how this is possible both ?

Kushal Bose
- 4 years ago

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@Kushal Bose Thank you for carefully reading the solution! I'll edit the solution. Ah I see, that is a typo!

Kazem Sepehrinia
- 4 years ago

Sir, I am a 11 class student and have some query regarding maths in general (i.e questions and ideas). Sir, I need to clarify my queries as I want to pursue this field further after 12 class. So sir, if you may can you provide some means of contact like Facebook, Twitter etc so that I may personally ask the doubts. Thank you, I have been a keen follower of your questions.

Sanjay Kumar
- 3 years, 12 months ago

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Hi, contact me at ksepehrinia@gmail.com

Kazem Sepehrinia
- 3 years, 12 months ago

why k=-1? why not k=-2 ... etc other negative terms?? it is nt said / suggested anywhere (a+b)=k=> has to be +ve integer or -1.

Ananya Aaniya
- 3 years, 12 months ago

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Square of a rational cannot be negative. Note that for $k \le -3$ the expression $\frac{2794-k}{3(k+2)}$ becomes negative and for $k=-2$ it is undefined.

Kazem Sepehrinia
- 3 years, 12 months ago

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If we write $n = a+b$ then we have $\begin{aligned} n^3 - 3abn + 3(n^2 - 2ab) & = 700n^2 \\ ab & = \frac{n^2(n-697)}{3(n+2)} \end{aligned}$ so that $a,b$ are the roots of the quadratic equation $X^2 - nX + \frac{n^2(n-697)}{3(n+2)} \; = \; 0$ Thus the discriminant of this quadratic must be the square of a rational, and hence $\frac{n^2(2794 - n)}{3(n+2)} \; = \; q^2$ for some rational $q$ . This means that $-2 < n \le 2794$ and that $\begin{aligned} (n+2)(2794 - n) & = 3k^2 \\ 5588 + 2792n - n^2 & = 3k^2 \\ (n - 1396)^2 + 3k^2 & = 1954404 = 2^2 \times 3^2 \times 233^2 \end{aligned}$ for some non-negative integer $k$ . Considering the factor of $3$ , it is clear that $n-1396$ , and hence also $k$ , is a multiple of $3$ . Thus $n = 1396 + 3u$ , $k = 3v$ where $u,v$ are integers with $v \ge 0$ and $u^2 + 3v^2 \; = \; 2^2 \times 233^2$ Since $233 \equiv 1 \pmod{4}$ we see that $\left(\frac{-3}{233}\right) \; = \; \left(\frac{-1}{233}\right) \left(\frac{3}{233}\right) \; = \; \left(\frac{233}{3}\right) \; = \; \left(\frac{2}{3}\right) \; = \; -1$ and so $-3$ is not a quadratic residue modulo $233$ . Thus $v$ , and so $u$ as well, is a multiple of $233$ . Thus we deduce that $n = 1396 + 699p$ , $k = 699q$ , where $p,q$ are integers with $q \ge 0$ and $p^2 + 3q^2 = 4$ . There are only four possible solutions, namely $(p,q) = (2,0), (-2,0), (1,1), (-1,1)$ , which yield $n = -2, 2794, 697, 2095$ respectively. Ignoring the first of these (since we know that $n > -2$ ), there are three possible values of $n$ , with one solution for $a,b$ for each such value, and so the answer to the question is $2794+697+2095 = \boxed{5586}$ .