$\frac{r^2}{s}+200=\frac{s^2}{r}+200^2$

In the equation above, $r$ and $s$ are both rational numbers and their difference is an integer.

What is the number of solutions to the equation?

The answer is 7.

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The equation becomes $r^3 - s^3 \; =\; N(N-1)rs$ where $N=200$ , and we want $r = s+n$ for some integer $n$ . Thus $\begin{aligned} 3s^2n + 3sn^2 + n^3 & = N(N-1)(s^2 + sn) \\ \big[N(N-1) - 3n\big]s^2 + n\big[N(N-1) - 3n\big]s - n^3 & = 0 \end{aligned}$ and so $s \; = \; -\tfrac12n \pm \frac{n}{2\big[N(N-1) - 3n\big]}\sqrt{\big(N(N-1)-3n\big)\big(N(N-1) + n\big)}$ Thus $-N(N-1) \le n < \tfrac13N(N-1)$ , and $\big(N(N-1) - 3n\big)\big(N(N-1) + n\big)$ must be a perfect square.

Simple calculations with $N=200$ shows that there are only $5$ values of $n$ for which this is true: $-39800, -39000, -6000, 0, 5200$ .

When $n=-39800$ , the only value of $s$ is $19900$ . When $n=0$ , the only value of $s$ is $0$ , and this is not a permitted solution of the original equation. The other three values of $n$ each yield two values of $s$ , and so there are a total of $\boxed{7}$ solutions.

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Could u explain me the calculations u did to get the 5 values for n?

Maninder Dhanauta
- 3 years, 11 months ago

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Set N= 200 into (N(N-1) - 3n) ( N(N-1) + n) = m^2 for some positive integer $m$ .

Expand and simplify, then apply Simon's favorite factoring trick .

Pi Han Goh
- 3 years, 11 months ago

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@Pi Han Goh - I get m^2 + 3n^2 +79600n =1584040000 but how do i apply that trick to this type of equation

Maninder Dhanauta
- 3 years, 11 months ago

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@Maninder Dhanauta – Sorry, let me start over: expand my original equation gives

-3n^2 - 79600n + 1584040000 = m^2

since we are looking for integer roots, then the quadratic discriminant of the equation above (in n) is a perfect square:

79600^2 - 4(-3)(1584040000 - m^2) = c^2, for some integer c

Now can you solve it from here?

Pi Han Goh
- 3 years, 11 months ago

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@Pi Han Goh – You've got a Pell equation. That opens a whole new bag of difficulties!

Kazem Sepehrinia
- 3 years, 11 months ago

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@Kazem Sepehrinia – No, we don't get Pell's equation.

We are looking to solve $\big(N(N-1) + n\big)\big(N(N-1) - 3n\big) \; = \; q^2$ for integers $n,q$ , where $N=200$ . Expanding and collecting terms, this becomes $\big[3n + N(N-1)\big]^2 + 3q^2 \; = \; 4N^2(N-1)^2$ From this it is clear that $n$ and $q$ must be of the same parity, so we can write $3n + N(N-1) = 2p-q$ and rewrite the condition in the form $\begin{aligned} (2p-q)^2 + 3q^2 & = 4N^2(N-1)^2 \\ p^2 - pq + q^2 & = N^2(N-1)^2 \\ (p + q\omega)(p + q\omega^2) & = N^2(N-1)^2 \end{aligned}$ working in the Euclidean domain $\mathbb{Z}[\omega]$ , where $\omega$ is the principal cube root of unity. Since $2$ and $5$ are both irreducible in this Euclidean domain, we deduce that $N = 200$ divides both $p$ and $q$ , and hence $p = NP, q = NQ$ where $(P + Q\omega)(P + Q\omega^2) \; = \; 199^2 \; = \; (15 + 2\omega)^2(15 + 2\omega^2)^2$ where the RHS is now a factorisation into irreducibles in $\mathbb{Z}[\omega]$ ,

It now follows that $P + Q\omega$ must be equal to one of $(15 + 2\omega)^2u$ , $(15 + 2\omega^2)^2u$ or $199u$ , where $u$ is one of the six units in $\mathbb{Z}[\omega]$ . Running through these $18$ cases, only a selection of them end up giving an integer value for $n$ . Wtthout giving the details, we obtain $-39000, -39800, -6000, 0, 5200$ as the only possible values of $n$ .

The way @Pi Han Goh was going with this, he would obtain an equation with $c^2 + 12m^2$ equal to a particular integer. This is basically the same as the one I have analysed, with an extra factor of $2$ to worry about.

This approach gives us a strategy for dealing with the general case of other values of $N$ . We have to factorise $N(N-1)$ into irreducibles in $\mathbb{Z}[\omega]$ ...

Mark Hennings
- 3 years, 11 months ago

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@Mark Hennings – On second viewing, I discovered that my work (without using Mark's Euclidean domain thingy technique, because I haven't mastered it yet!!) still requires plenty of trial and error. But I did try to minimize the amount of trials needed. Here's my failed attempt:

From what I've written, we find to find non-negative integer solutions to $c^2 + 12m^2 = 25344640000$ .

Consider modulo 4, we can see that $c^2 \bmod 4= 0$ . Thus, let $c = 2c_1$ and $m =m_1$ (for consistency sake). Then the equation simplifies to $c_1 ^2 + 3m_1 ^2 =6336160000$ .

Consider modulo 5. The quadratic residues of 5 are 0,1 and 4. Since RHS is divisible by 5, then so does LHS. Thus $c_1$ and $m_1$ must be divisible by 5. Let $c_1 = 5c_2, m_1 = 5m_2$ . The equation simplifies to $c_2^2 + 3 m_2 ^2 = 253446400$ .

Repeat the process again: Consider modulo 5. Since RHS is divisible by 5, then so does LHS. Thus $c_2$ and $m_2$ must be divisible by 5. Let $c_2 = 5c_3, m_2 = 5m_3$ . The equation simplifies to $c_3^2 + 3 m_3 ^2 = 10137856 = 3184^2$ .

Not sure how to finish this off (elegantly)... I was thinking of sum of 2 (or 4) squares theorem, but I'm pretty sure that will fail.

Pi Han Goh
- 3 years, 11 months ago

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Let $A=200^2-200$ . Thus $r^3-s^3=Ars$ . Write $r-s=n$ , where $n \in \mathbb{Z}$ . Then $(s+n)^3-s^3=A(s+n)s$ , or $(A-3n)s^2+(An-3n^2)s-n^3=0$ According to quadratic formula, we get $s=-\frac{n}{2}\pm \frac{|n|}{2} \frac{\sqrt{(A+n)(A-3n)}}{A-3n}$ Therefore, $(A+n)(A-3n)=k^2$ , where $k$ is a non-negative integer. It follows that $-A \le n \le \lfloor A/3 \rfloor$ .

Note that $A$ is divisible by $200$ . In the equation $(A+n)(A-3n)=k^2$ look mod $5$ , then $-3n^2 \equiv k^2 \pmod{5}$ . Hence $n \equiv k \equiv 0 \pmod{5}$ , since a perfect square mod $5$ is congruent to $0, 1, 4$ only. If you divide both sides of the equation by $25$ and look mod $5$ again, then it comes out that $n \equiv k \equiv 0 \pmod{25}$ .

In the equation $(A+n)(A-3n)=k^2$ look mod $8$ , then $-3n^2 \equiv k^2 \pmod{8}$ . Hence $n$ and $k$ are both even, since a perfect square mod $8$ is congruent to $0, 1, 4$ only. If you divide both sides of the equation by $4$ and look mod $8$ again, then a similar argument gives $n \equiv k \equiv 0 \pmod{4}$ . Now write $n=4n'$ and $k=4k'$ and divide both sides of the equation by $16$ to get $(A/4+n')(A/4-3n')=k'^2$ . Look mod $8$ , then $4+4n'+5n'^2 \equiv k'^2 \pmod{8}$ . If $n'$ is odd, then LHS of this congruence is $5$ mod $8$ , which is contradiction. Thus $n'$ and $k'$ are both even and finally $n \equiv k \equiv 0 \pmod{8}$ .

Thus in total we have that $n \equiv k \equiv 0 \pmod{200}$ . Write $n=200 N$ and $k=200 K$ to get $(199+N)(199-3N)=K^2$ For $\color{#D61F06}N=-199$ and $\color{#D61F06}N=0$ the LHS becomes a square. Then, for $N \not\equiv 0 \pmod {199}$ we have that $d=\gcd(199+N, 199-3N) \mid 3(199+N)+199-3n=4\times 199$ Hence $d=1, 2, 4$ .

If $d=1$ , then $199+N=a^2$ and $199-3N=b^2$ , where $a$ and $b$ are co-prime odd positive integers. Hence $3a^2+b^2=796$ . Its easy to see that $a \le 15$ . Checking $a=1, 3, 5, ..., 15$ , reveals that $a=13, 15$ make $796-3a^2$ a perfect square. Thus we get two pairs: $\color{#3D99F6}(a, b)=(13, 17), (15, 11)$ that result in $\color{#D61F06}N=-30$ and $\color{#D61F06}N=26$ , respectively.

If $d=2$ , then $199+N=2a^2$ and $199-3N=2b^2$ , where $a$ and $b$ are co-prime positive integers. It follows that $3a^2+b^2=398$ . Its easy to see that $a \le 11$ . Check that none of $a \le 11$ make $398-3a^2$ a perfect square.

If $d=4$ , then $199+N=4a^2$ and $199-3N=4b^2$ , where $a$ and $b$ are co-prime positive integers. It follows that $3a^2+b^2=199$ . Its easy to see that $a \le 7$ . Checking $a \le 7$ , reveals that $a=1$ makes $199-3a^2$ a perfect square. Thus we get one pair: $\color{#3D99F6}(a, b)=(1, 14)$ that results in $\color{#D61F06}N=-198$ .

Finally we get $5$ solutions for $n=200N$ , namely $n=-39800, -39000, -6000, 0, 5200$ $n=-39800$ gives only $s=19900$ . $n=0$ gives $A=0$ , which is a contradiction! Other values of $n$ give two different solutions for $s$ . Thus the total number of solutions for the original equation is $\boxed{7}$ .

Notice that if $(r, s)$ is a solution, then $(-s, -r)$ is a solution too. Thus I write four of the solutions to the original equation. $(r, s)=(s+n, s)=\left(-19900, 19900 \right), \left(-\frac{126750}{7}, \frac{146250}{7} \right), \left(-\frac{12000}{17}, \frac{90000}{17} \right), \left( \frac{67600}{11}, \frac{10400}{11} \right)$