Find all solutions of the equation $y^4+6y^2 = 7z^2+12$.

Question

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.

Answer 1

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

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 $(y,z)=( 2, 2).$

Answer 2

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