Out of the infinitude of integers >= 0, how does one go about proving ...?

Question

... that there are no solutions to y^2 = 3x^3 + 6x^2 + 6x + 16 other than (x,y) = (0,0) for integral x and y? Well, that's what I surmise! I'd like to be able to have a general method of attack, even for much simpler equations such as y^2 = 8x + 7, which doesn't seem to have any solutions at all.

Sorry geezerbill and others. I meant to write the only solution as (x,y) = (0,4).

Answer 1

Actually, plugging in (0,0) into your first equation gives 0 = 16, so that doesn't even work.

As for y² = 8x +7, I suppose there are a couple of ways to prove there are no solutions. Note that you can rewrite this as
y² = 8x+(8-1)
y² = 8(x+1)-1
This expression on the right is always going to be an odd number, so y² and thus y too both have to be odd. An odd number must end in 1, 3, 5, 7 or 9, so the square can only end in 1, 5 or 9.

If y ends in 5, then y² ends in 25 (this is something you can prove on the side). So 8(x+1) has to end in 26. Note however that a number ending in 26 can't be divisible by 4, using the old divisibility rule for 4. Thus it can't be divisible by 8 either. So y² must end in 1 or 9.

Consider what happens when you take a number ending in 1, 3, 7, or 9 and square it. You can always write the first number as multiple of 10 plus the last digit (e.g., 349 = 34*10 + 9) . Here's what you get when you square these numbers:

(10n+1)² = 100n² + 20n + 1
(10n+3)² = 100n² + 60n + 9
(10n+7)² = 100n² + 140n + 49
(10n+9)² = 100n² + 180n + 81

If you think of these as all the possible forms of y², then adding 1 to each gives the possible forms of 8(x+1). Add 1 to each one, and you can rewrite them as:

4(25n + 5n) + 2
4(25n + 15n) + 10
4(25n + 35n) + 50
4(25n + 45n) + 82

But none of these values are divisible by 4, because they're all multiples of 4 plus a number that isn't divisible by 4. Therefore, none of them can be multiples of 8 either. So there are no integral solutions to y² = 8x + 7.

Hopefully this will give you some ideas on how you might try to tackle the larger problem. I imagine though that there are some other tricks, using a lot of modular arithmetic for example.

Answer 2

This is a good question. Solutions where the variables have to be integers are called Diophantine equations. There is no general method for solving them, but there are some useful techniques. One is to divide possible solutions into different classes and try each one out.

For example, in the case of y² = 8x + 7, it is obvious that y² is odd, so y must be odd. Any odd number must have a remainder of 1 or 3 when divided by four. So it must either be of the form 4k+1 or 4k+3 for some number k.

Let's try these out:

y = 4k + 1
y² = 16k² + 8k + 1
= 4(4k² +2k) + 1
= 4m + 1 for some m.

So y² is 1 greater than a multiple of 4. But 8x + 7 is not. It is 3 greater than a multiple of 4. So a y of this form (4k + 1) can not satisfy the equation.

Now try y = 4k +3

y=4k + 3
y² = 16k² + 24k + 9
= 16k² + 24k + 8 + 1
= 4(4k² + 6k + 2) + 1
= 4m + 1 for some m

So again y² is 1 greater than a multiple of 4, so it can't be equal to 8x + 7, which is 3 greater than a multiple of 4.

Therefore y² = 8x + 7 has no integer solutions. You could try adapting this approach to the more difficult problem.

Answer 3

Ok this is a Diophantine equation, which in general do not have an algebraic solution. I am not aware of a method to disprove the existence of solutions.

The RHS bears a strong resemblance to 3(x+1)³
Let's rewrite it slightly:
y² = 3x³ + 6x² + 6x + 16
y² = 3(x³ + 2x² + 2x + 1) + 13
y² = 3(x³ + 3x² + 3x + 1) - 3(x² +x) + 13
y² = 3(x+1)³ - 3x(x+1) + 13
y² = 3(x+1)[(x+1)² - x] + 13
y² = 3(x+1)(x² + x +1) + 13

writing w = x+1
y² = 3w³ -3w(w-1) + 13
or
y² = 3w(w² -w+1) +13


maybe this helps e.g. noting the RHS always has modulus
(13 modulo w)

I think one line of attack is to consider the residues modulo some number or power of two (e.g. 4,8). This then gives you a restriction on possible solutions.

So let's try modulo 13, writing w ≡ a (mod 13)
i.e. w = 13k + a:

y² = 3w(w² -w+1) +13
y² ≡ 3a(a² -a+1) + 0 (mod 13)
May be helpful.

Trying w = 4k + a
y² = 3w(w² -w+1) +13
y² ≡ 3a(a² -a+1) +1 (mod 4)
So:
a=0 => y² ≡ 1 (mod 4)
a=1 => y² ≡ 0 (mod 4)
a=2 => y² ≡ 3 (mod 4) IMPOSSIBLE
a=3 => y² ≡ 1 (mod 4)

As Gnomon notes, integer squares (y²) can only ≡ 0 or 1 mod 4

Trying w = 8k + a
y² = 3w(w² -w+1) +13
y² ≡ 3a(a² -a+1) +5 (mod 8)

So:
a=0 => y² ≡ 5 (mod 8) IMPOSSIBLE
a=1 => y² ≡ 0 (mod 8)
a=2 => y² ≡ 7 (mod 8) IMPOSSIBLE
a=3 => y² ≡ 4 (mod 8)
a=4 => y² ≡ 1 (mod 8)
a=5 => y² ≡ 0 (mod 8)
a=6 => y² ≡ 3 (mod 8) IMPOSSIBLE
a=7 => y² ≡ 4 (mod 8)

Again, integer squares (y²) can only ≡ 0,1,4 mod 8

The only other way I can see to come at it is to write y
module some base j. Noting that any factors of y must be double-factors of y², then you investigate what the implications of that are for the RHS and try to prove a contradiction.
In particular I presume you focus on some prime factor(s) of y:
y = mj+b ≡ b (mod j)
y² ≡ b² (mod j)

y² = 3w(w² -w+1) +13