1

Find all positive integer solutions to the equation $y^4+6y^2 = 7z^2+12$

So far I have tried to take different modulos of the equation like 5, 6, and tried to find a relationship between y and z. However, it gets complicated with multiple cases very quickly. How would one narrow down the number of cases?

Also, the only solution to this equation is y = z = 2.

Oshawott
  • 3,956
Prathmesh
  • 91

2 Answers2

4

Hint: We can rewrite the equation as $$ (y^2+3)^2=7(z^2+3). $$ Now consider primes $p$ dividing one side.

Dietrich Burde
  • 130,978
  • Okay. Taking mod 7 gives that y has to be either 2 or 5 mod 7. Lets just consider the 2 mod 7 case for a second. Let y = 7y_1 + 2. Plug it in and you get a whole mess of equations.

    $2401 y^4 + 2744 y^3 + 1470y^2 +392y + 49 = z^2+3$ Similarly, z also has to be 2 or 5 mod 7. Where do I go from here. It seems to get really messy.

     – Prathmesh Jul 11 '21 at 23:15
  • Note that then $(y^2+3)^2$ is divisible by $7^2$, so $z^2+3$ is divisible by $7$. Also, without loss of generality $y\equiv z\equiv2\pmod{7}$ by changing signs. – Servaes Jul 13 '21 at 21:11
  • That doesn't really help though. Because sure you can keep going and then by infinite descent one can show that y = z = 2. But what about when y and or z and their smaller numbers are -2 mod 7 or 5 mod 7. Also, what if it alternates? What do we do then?

    @Dietrich Burde Please post a full solution if you have one...

     – Prathmesh Jul 21 '21 at 19:35
  • A full solution is already posted by Tomita, and you have accepted the answer. Where does the question come from? Are you assumed to use certain methods? – Dietrich Burde Jul 25 '21 at 14:39
3

$$7y^4+42y^2 = 49z^2+84\tag{1}$$

Let $U=y, V=7z$ then we get

$$V^2 = 7U^4+42U^2-84\tag{2}$$

An equation $(2)$ is birationally equivalent to the elliptic curve
$$Y^2 = X^3 + 1764X + 71344 \tag{3}$$
with $$U = \frac{1176-2Y}{-Y+14X-196}, V = \frac{16464Y-2352X^2-3380608+14X^3-24696X}{(-Y+14X-196)^2}$$

According to LMFDB , elliptic curve has five integer points as follows.

$(X,Y)=(-28, 0) , (21,\pm 343), (56,\pm 588).$

From $(X,Y)=(-28, 0) \implies (U,V)=(-2, -14) \implies (y,z)=(-2,-2).$

From $(X,Y)=(21, 343) \implies (U,V)=(-2, 14) \implies (y,z)=(-2, 2).$

Thus, there is only positive integer solution $(y,z)=( 2, 2).$

Tomita
  • 2,346
  • I have never worked with elliptical curves. Is there any way to do this question without using those? If not, where can I get a quick breakdown of the concepts you used in solving this problem? – Prathmesh Jul 12 '21 at 13:00
  • 1
    @Prathmesh To the best of my knowledge, I don't know of any other easy way. I recommend the book "Rational Points on Elliptic Curves" by Silverman-Tate. I like this book. About "Transform quartic curve to elliptic curve", pls see https://math.stackexchange.com/questions/1591990/birational-equivalence-of-diophantine-equations-and-elliptic-curves About "integer points of elliptic curve", https://mathoverflow.net/questions/6676/integer-points-of-an-elliptic-curve – Tomita Jul 12 '21 at 23:48