Finding a Square Sum

It is necessary to know that all perfect squares are congruent to $0$ or $1$ $(mod 4)$ , i.e.

$x^2 \cong 0$ or $1 (mod 4) \forall x \in Z$ .

Now, we see that $y$ can be written as either one of: $(4n + 1), (4n + 2), (4n + 3)$ or $(4n + 4)$ for some $n \in Z^{+}$ . Considering each case:

$Case 1: y = 4n + 1$ for some $n \in Z$ $y^3 + 7 = (4n + 1)^3 + 7 = 64n^3 + 48n^2 + 12n + 8 = 4(16n^3 + 12n^2 + 3n + 2)$ . This expression cannot be factored because of the Rational Root Theorem and hence cannot be a perfect square for any $x \in Z$ , since $n$ is an integer.

$Case 2: y = 4n + 2$ for some $n \in Z$ $y^3 + 7 = (4n + 2)^3 + 7 = 64n^3 + 96n^2 + 48n + 11 \cong 3 (mod 4)$ ; hence, this expression cannot be equal to $x^2$ for $x \in Z$ .

$Case 3: y = 4n + 3$ for some $n \in Z$ $y^3 + 7 = (4n + 3)^3 + 7 = 64n^3 + 144n^2 + 108n + 34 \cong 2 (mod 4)$ ; hence, this expression cannot be equal to $x^2$ for $x \in Z$ .

$Case 24: y = 4n + 4$ for some $n \in Z$ $y^3 + 7 = (4n + 4)^3 + 7 = 4^3(n + 1)^3 + 7 \cong 3 (mod 4)$ ; hence, this expression also cannot be equal to $x^2$ for $x \in Z$ .

Hence, there are $\boxed {0}$ integer solutions $(x, y)$ such that $x^2 = y^3 + 7$ .

Working in the system $\mathbb{Z}[\sqrt{7}]$ , we can rewrite what we have as $(x-\sqrt{7})(x+\sqrt{7})=y^3$ . Since $y^3$ can be expressed as a number times its conjugate, $y$ must split into $(a+b\sqrt{7})(a-b\sqrt{7})$ where $a,b\in\mathbb{Z}$ . So then $y^3=(a+b\sqrt{7})^3(a-b\sqrt{7})^3$ . So either $(a+b\sqrt{7})^3=(x-\sqrt{7})$ or $(a-b\sqrt{7})^3=(x-\sqrt{7})$ . We can consider the two cases.

Case 1: $(a+b\sqrt{7})^3=(x-\sqrt{7})$ Then we have $(a^3+21ab^2)+(3a^2b+7b^3)\sqrt{7}=x-\sqrt{7}$ . So that means $3a^2b+7b^3=-1$ , but there are no solutions $(a,b)$ which satisfy this because the absolute value of that quantity would be greater than 1.

Case 2: $(a+b\sqrt{7})^3=(x+\sqrt{7})$ This is similar to the first case. We get $(a^3+21ab^2)+(3a^2b+7b^3)\sqrt{7}=x-\sqrt{7}$ , so $3a^2b+7b^3=1$ and again, there are no solutions.

[Note: This solution uses several (unjustified) facts from field theory]

Consider $x^2 = y^3 + 7 \pmod{4}$ .

Case 1: If $y$ is even, then $x^2 \equiv 3 \pmod{4}$ , which is a contradiction since 3 is not a quadratic residue modulo 4. Hence, we have that $y$ is odd and $x$ is even.

Case 2: If $y \equiv 3 \pmod{4}.$ , then $x^2 = y^3 + 7 \equiv 2 \pmod{4},$ which is a contradiction.

Case 3: If $y \equiv 1 \pmod{4}$ . Since $x^2 + 1 = (y+2)(y^2 - 2y +4),$ it is convenient to look at the divisors of $x^2 + 1.$ Suppose it has a prime factor of the form $p = 4k +3.$ Then $x^2 \equiv - 1 \pmod{p}$ , so $x^{4k+2} \equiv (-1)^{2k+1} = - 1 \pmod{p}$ , contradicting Fermat's little theorem. So all of the prime divisors of $x^2 + 1$ are congruent to 1 mod 4. If $y \equiv 1 \pmod{4}$ , then $y+2 \equiv 3 \pmod{4}$ , and must certainly have a prime factor congruent to 3 mod 4. (Indeed, or it would be 1 mod 4 otherwise.) But $y+2|x^2 + 1$ , which contradicts the fact that $x^2 + 1$ has all prime factors 1 mod 4.

Hence, we have no solutions for $y$ , giving the answer $0$ .

Suppose on the contrary that there are integers solution, We will first look at $y$ being even. This will mean that that $y^3 \equiv 0 \pmod{8}$ , $y^3+7 \equiv 7 \pmod{8}$ , $x^2 \equiv 7 \pmod{8}$ . This is not possible, as $a^2 \equiv 0,1,4 \pmod{8}$ is the only possible quadratic residues. Hence a contradiction, therefore $y$ is not even.

Now for an odd $y$ , we consider $x^2+1=y^3+8=(y+2)(y^2-2y+4)$ . Now, $y^2-2y+4=(y-1)^2+3$ which is always positive. Since $y$ is odd, $y-1$ is even and hence $(y-1)^2 \equiv 0 \pmod{4}$ , so $(y-1)^2+3 \equiv 3 \pmod{4}$ So there must be a prime number, $p$ , that divides $(y-1)^2+3$ with the property that $p \equiv 3 \pmod{4}$ . So $x^2+1$ must divide $p$ also or $x^2+1 \equiv 0 \pmod{p}$ $x^2 \equiv -1 \pmod{p}$ . But then again this is not possible as $p$ is in the form $p=4n+3$ , which implies that $(-1)^{\frac{p-1}{2}} \equiv -1 \neq 1 \pmod{p}$ . This means that $x^2 \equiv -1 \pmod{p}$ is not possible by Euler's Criterion on Quadratic Residues. Hence there can be no integer solutions.

$x^2 = y^3+7$

$x^2+1=y^3+8=(y+2)(y^2-2y+4)$

It is pretty apparent that $y > -2$ , or else $x^2$ would be negative.

If $y$ is even then,

x^2  \equiv 7 \equiv -1\pmod{4}

But, it is well-known that the remainder of a square can only be $0$ or $1$ . Hence, $y$ cannot be even. Thus, assume $y$ is odd. Since $y$ is odd, $x$ must be even.

Also, given the fact that $y$ is odd,

$y^3 \equiv y \pmod{4}$

So, $x^2 \equiv y^3+7 \pmod{4}$ is equivalent to

$0 \equiv y -1 \pmod{4}$

Then, $y+2 \equiv 3 \pmod{4}$ .

Since $y+2$ is positive, odd and congurent to $3$ mod $4$ , it must have a prime factor $p$ such that

$p \equiv 3 \pmod{4}$ .

Since $p|y+2$ , and $y+2|x^2+1$ ,

$y^2+1 \equiv 0 \pmod{p}$

Or,

$y^2 \equiv -1 \pmod{p}$ ,

which is only solvable if $p \equiv 1 \pmod{4}$ , which is not the case.

Thus, the given equation has no integral solutions.

Let (x,y) be a solution. Add one to both sides, obtaining $(x^2+1) = y^3+8$ , or $x^2+1 = (y+2)(y^2-2y+4)$ . If y was even, $x^2 \equiv7 \pmod 8$ , but 7 is not a quadratic resiude mod 8. Thus y is odd. This implies that $y^2-2y+4 \equiv y(y-2) \equiv3 \pmod 4$ . This gives a contradiction, because it is well known that all odd factors of $k^2+1$ are 1 mod 4. Thus there are 0 solutions to the equation.

$x^2 = y^3 + 7 \to y^3 = x^2 - 7$

$y^3 = (x + \sqrt{7})(x - \sqrt{7})$

Therefore $(x+\sqrt{7})^2 = (x - \sqrt{7})$ or $(x-\sqrt{7})^2 = (x + \sqrt{7})$

or either one of $(x + \sqrt{7})$ and $(x - \sqrt{7}) = 1$

As $x$ is an integer, the above condition is impossible.

Considering the two equations in the first case,

i. $x^2 - (2\sqrt{7} + 1)x + (7 - \sqrt{7}) = 0$

ii. $x^2 + (2\sqrt{7} - 1)x + (7 + \sqrt{7}) = 0$

None of these have integer solutions so there are 0 pairs of $(x,y)$ .

Negative numbers and zero cannot be a solution for y since x^2 becomes negative for y<-1 while y=-1,0 does not satisfy the equation. For positive numbers, it can be observed that the differences (Note: x^2-y^3=7) between squares (nearest the value of the cube) and cubes are significantly greater than 7, and this becomes more apparent in larger values of x and y.

x^2=y^3+7, x^2+1^2=y^3+2^3, Sum of two integer squares cannot be equal to sum of two integer cubes(only exception is combination of 0 and 1, 0^2+1^2=0^3+1^3) .Here already 2 cube is present in the rhs ,so that no possibility exist