Prove that the equation $y^2=x^3-73$ has no integer solutions Prove that there are no integers $x,y$ such that  $y^2=x^3-73$.
Thank you.
 A: Equations of the form $$ y^2 = x^3 + k$$ are known as Bachet equations. I will quote the statement of Theorem 4.2 from Richard Mollin's Algebraic Number Theory:

Let $F=\mathbb{Q}(\sqrt{k})$ be a complex quadratic field with
  radicand $k< -1$ such that $k \neq 1 \pmod 4,$ and $h_{\mathfrak{D}_F}
 \neq 0\pmod 3.$ Then there are no solutions to the Batchet equation in
  integers $x,y$ except in the following cases: there exists an integer
   $u$ such that $$(k,x,y) = (\pm 1-3u^2, 4u^2 \mp 1, \epsilon \cdot u(3
 \mp 8u^2) ),$$
where the $\pm$ signs correspond to the $\mp$ signs and $\epsilon =
 \pm 1$ is allowed in either case.

A: You should send the equation into the congruence modulo $4$, for that according to Division Algorithm, (or Complete Residue System modulo 4) we have 4 possibilities for arbitrary number $x$.
$$ \text{If}  \  x \equiv 0 \ \text{or} \ 2 \pmod{4} \ \ \text{then} \ \ y^2\equiv 0 - 73 \equiv 3 \equiv -1 \pmod{4} $$
So as we know in the congruence modulo $4$, it's not possible for a squared number (a number raised to the power of $2$) to be $-1$, all squared numbers are congruent to either $1$ or $0$ modulo $4$.
$$\text{If} \ \ x\equiv 3 \pmod{4}\ \ \text{then} \ \  y^2\equiv -1 -73\equiv2 \pmod{4} \ \ $$
And again it's not possible. The last part is $ x\equiv1\pmod{4} $ which is a little tricky! Look upon our equation and increase both side of it by $100$, the new equation will be: $$y^2+100=x^3+27 $$ By some algebraic factorising reformulate it like this: 
$$y^2+100=(x+3)(x^2-3x+9)\qquad (1)$$
If we take the single factor $(x^2-3x+9)$ and notice that $$x^2-3x+9 \equiv 1-3+9\equiv-1 \pmod{4}$$ 
Our game is started from now on, there must be an odd prime like $q$ in which, $q\equiv -1 \pmod{4}$ and $q \mid x^2-3x+9$.
Actually that's very convenient and there is an odd prime like $q$ because if we show $x^2-3x+9$ as the product of $p_1,p_2,...,p_k $ (note that it's possible for a pair or more than two of them like $p_i$ and $p_j$ to have $p_i=p_j$ and this will cover $p_1^\alpha p_2^\beta ...$ the true Unique-Prime-Factorization representation of numbers according to Fundamental Theorem of Arithmetic) and neither every $p_i$ can be congruent to $1$ modulo $4$ nor the number of primes which are congruent to $-1$ is even. 
We show that the second part must be true, first we know every odd prime should be congruent to either $1$ or $-1$ modulo $4$ and also:
$$ x^2-3x+9 = p_1p_2...p_k \equiv 1\times 1\times ... (-1)\times (-1)\times ...\equiv (-1)^n \pmod{4}$$ So it's obvious for $n$ to be odd. Let's get back to our found $q$ and our customized equation (1). 
$$ y^2+100\equiv (x+3)(x^2-3x+9) \equiv 0 \pmod{q}$$ 
Therefore:
$$y^2 \equiv -100 \pmod{q}$$
$100=2^2\times 5^2$ and both $2$ and $5$ are primes and their remainders when they're divided by 4 are 2 and 1. So $2 \nmid q, q\nmid 5$ or better say $\text{gcd}(5,q)=\text{gcd}(2,q)=\text{gcd}(2\times 5,q)=1$, therefore there is an inverse of $10$ module $q$ and let denote it with $10\ '$, and multiply both sides of the congruence by $(10\ ')^2$:
$$ (10\ ')^2y^2\equiv -1 \pmod{q} $$
And we turn $10\ 'y$ into a new variable like $z$, therefore:
$$z^2\equiv -1 \pmod{q}$$ This congruence has solution if and only if $q\equiv 1 \pmod{4}$ (according to a theorem which is resulted in by famous Wilson's Theorem), but our poor $q$ is congruent to $-1$ modulo $4$, and eventually we proved that  $y^2=x^3-73$ has no solution among integers.
