Let p be a prime. Prove that $x^4 + 4 y^4 = p$ has integer solutions if and only if $p = 5$. In this case, find the solutions for x and y.

Question

I have tried to factor

$$x^4 + 4 y^4 = p$$

using the Sophie Germain's Identity, as someone suggested in the comments, which yields:

$$((x+y)^2 + y^2)((x-y)^2 + y^2) = p$$

Since p is a prime, either

$$(x+y)^2 + y^2 = 1$$

or

$$(x-y)^2 + y^2 = 1$$

I do not know how to solve this for $x,y$. And I do not see how this will proves that p can only be 5. Furthermore, I do not know how to find the integer solutions for the equation:

$$x^4 + 4 y^4 = 5$$

Answer 1

Here's a hint: $$x^{4} + 4y^{4} = (x^2)^2 + (2y^{2})^{2} + 4x^{2}y^{2} - 4x^{2}y^{2} = (x^{2} + 2y^{2})^{2} - (2xy)^{2}.$$ Can you factor this further?

If $x^{4} + 4y^{4} = p$ is prime, what does this say about the factors of $x^{4} + 4y^{2}$?

(Edited to give more details)

So, we have $$x^{4} + 4y^{4} = (x^{2} + 2y^{2})^{2} - (2xy)^{2}.$$ Since this is a difference of squares, we can factor it as $$(x^{2} + 2y^{2}- 2xy)(x^{2}+2y^{2}+2xy).$$

Note that $$x^{2} +2y^{2} -2xy = (x^{2}-2xy+y^{2})+y^{2} = (x-y)^{2} + y^{2},$$ and that $$(x^{2}+2y^{2}+2xy) = (x^{2}+2xy+y^{2})+y^{2} = (x+y)^{2} + y^{2}.$$ Both of these are sums of squares, hence are nonnegative.

So, $x^{2} +2y^{2} -2xy$ and $x^{2}+2y^{2}+2xy$ are nonnegative integers which divide the prime $x^{4} +4y^{4} = p.$ So, we know that one of them must be $1.$

Case 1: Suppose that $x^{2} +2y^{2} -2xy = 1.$ Then, we have $(x-y)^{2} + y^{2} = 1.$ Since $(x-y)^{2}, y^{2}$ are both nonnegative integers, we must either have $(x-y)^{2} = 1$ and $y^{2} = 0,$ or $(x-y)^{2} = 0$ and $y^{2} = 1.$

The first possibility results in the possible solutions $(x, y) = (1, 0), (-1, 0)$, and the second possibility results in the possible solutions $(x, y) = (1, 1), (-1, -1).$

Note that if we substitute $(x, y) = (1, 0)$ or $(x, y) = (-1, 0)$ into $x^4 +4y^{4},$ we get $x^{4} + 4y^{4} = 1,$ which is not a prime. So, we eliminate those two solutions from consideration. On the other hand, if we substitute $(x, y) = (1, 1)$ and $(x, y) = (-1, -1)$, we get $x^{4} + y^{4} = 5.$ So, in the first case, we must have $p = 5,$ and we have the corresponding solutions $(x, y) = (1, 1), (-1, -1).$

Case 2: Now, suppose that $x^{2} +2y^{2} + 2xy = 1.$ Then, we get $(x+y)^{2} + y^{2} = 1.$ Using the same argument as in Case $1$, you get the possible solutions $(x, y) = (1, 0), (-1, 0), (-1, 1), (1, -1).$ (You should work out the details here for yourself!) The first two possibilities were already eliminated. If you substitute the remaining two $(x, y) = (-1, 1), (1, -1)$ into the equation, you get $$x^{4} +4y^{4} = 5.$$ So, in the second case, we must also have $p =5,$ and we have the corresponding solutions $(x, y) = (1, -1), (-1, 1).$

To conclude, we note that the two cases we discussed above are all the possibilities (since one of $x^{2} +2y^{2} -2xy$ and $x^{2}+2y^{2}+2xy$ must be $1$). Therefore, we must have $p =5,$ and we have the solutions $$(x, y) = (1, 1), (-1, -1), (1, -1), (-1, 1).$$

Answer 2

The only prime ramified in the ring of Gauss integers $\mathbb Z[i]$ it is known to be $2=i^3((1+i)^2$ so we have since $i^4=1$ $$x^4+4y^4=(x^2+i2y^2)(x^2-i2y^2)=(x^2+(1+i)^2y^2)(x^2-(1+i)^2y^2)$$ then the factorization in $\mathbb Z[i]$ of $x^4+4y^4$ becomes $$(x+i(1+i)y)(x-i(1+i)y)(x+(1+i)y)(x-(1+i)y)$$ which gives $$(x-y+iy)(x+y-iy)(x+y+iy)(x-y-iy)=[(x-y)^2+y^2][(x+y)^2+y^2]$$ Consequently we must consider the two cases $(x-y)^2+y^2=1$ and $(x+y)^2+y^2=p$ and the correlative $(x+y)^2+y^2=1$ and $(x-y)^2+y^2=p$ which easily give the solution to the problem.