# 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.

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$(2a+1)^2+73=2(2a^2+2a+37)$ even but not divisible by $8,$ so $y$ must be even. –  lab bhattacharjee Dec 10 '12 at 10:33
Equations of the form $y^2=x^3-k$ can sometimes be solved by purely elementary considerations, sometimes by factorization in the field with $\sqrt{-k}$, sometimes by factorization in the field with $\root3\of k$, and sometimes more advanced methods are needed. That's one reason it would be helpful to know the context in which you encountered the problem. –  Gerry Myerson Dec 10 '12 at 12:10
Why did this question get 5 upvotes? Bad manners, no context or trace of effort on the poster's part. –  Stefan Smith Apr 30 '13 at 22:43

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.
Do you know whether this field has class number divisible by $3$? If it does, the quote from Mollin doesn't help very much. Also, the equation is often referred to as the Mordell equation. –  Gerry Myerson Dec 10 '12 at 12:06