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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
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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
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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

2 Answers 2

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.

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Also known as Mordell's equation. –  lhf Dec 10 '12 at 12:03
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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
    
@GerryMyerson Mathematica computes the class number to be 4. –  Ragib Zaman Dec 10 '12 at 12:31

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: $$y^2+100=x^3+27 $$ And from high school we know algebraic factorising we write 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 not every $p_i$ can be congruent to $1$ modulo $4$ nor the number of primes which are congruent to $-1$ be 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 show 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.

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