Can X^4+nY^4 Always Produce a Prime Number?

[robert Ihnot]

Let x, y, n all represent positive integers in x^4+nY^4. It seem there is a lot of primes in this set. In fact, even allowing x=1, n=1, we look at 1+Y^4, we see pairs, y=1, f(y)=2, (2,17), (4,257), (6,1297), (16,65537), (20,160001) Possibly an infinite set?

Take the case of x=1, n=2, giving (1,3) now we have a problem since the form 1+2Y^4, will be divisible by 3 unless we take y as a multiple of 3 giving primes: (3,163), (6, 2593), (18,209953). In the case of x=1, n=3, 1^4+3*4^4 =769, (6, 3889),(8,12289)

We can continue with this, increasing n, but in the case of x^4+4y^4 there is only one prime solution, x=1, y=1, F(x,y) = 5.

QUESTION: Is there an n such that X^4+nY^4 NEVER gives a prime number?

 

[disregardthat]

Yes, take $n = 64=4 \cdot 2^4$. Then $$x^4+ny^4=x^4+4(2y)^4=(x^2+4xy+8y^2)(x^2-4xy+8y^2)$$. This is number is composite, since by the AM-GM-inequality $$x^2+8y^2-4xy \geq 2\sqrt{x^2 \cdot 8y^2}-4xy=(\sqrt{2}-1)4xy \geq (\sqrt{2}-1)4>1$$.

It can easily be seen that if $n = 4k^4$ for any integer k>1, $x^2+ny^4$ will never be prime. Now the question remains whether 64 is the least number...

 

[robert Ihnot]

They split up in the form, 2^2, 2^6, 2^10...the Y term then is 4y^4, 4(2y)^4, 4((4y)^4)...There seems no other factorization(?).

If we have ( x^2+axy+by^2)(x^2-axy+by^2) this gives x^4+(b^2)y^4 with a middle term of (2b-a^2)(xy)^2. to set that 0, obviously both a and b are divisible by 2. The simplist answer then is b=2. Otherwise the answer is b=2c^2, a=2c. This results in the term on Y is (bc^2)^2=b^2(c^4). c=1, gives 4, c=2 give 64, c=3 gives 324.

X^4+324y^4 = (x^2-6xy+18y^2)(x^2-6xy+18y^2).