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I have some trivial issues that are imposing me from understanding a proof from a book.

(i) if $y^2=8k^3+7$ where $k\in \mathbb{Z}$ $\implies$ $y^2\equiv7 \,[8]$ which is not possible, because $(\frac{7}{8})=-1$. But can someone tell me an easy way to interpret why $(\frac{7}{8})=-1$?

(ii) if $x$ is odd then I know why $(x-1)^2+3\equiv3 \,[4]$, but why does this imply that exists prime $p$ st $p|(x-1)^2+3\equiv3 \, [4]$ where $p\equiv3 \, [4]$? (according to book, otherwise all its prime factors are 1 modulo 4 $\implies$ $(x-1)^2+3\equiv1 \, [4]$) fair enough, but why does this $\implies$ $(x-1)^2+3\not\equiv3 \, [4]$?

can a number not be congruent 1 [4] and 3 [4] simultaneously?

And does it have to be congruent to at least one of these?

(iii) $p\equiv3 \, [4]$ $\implies$ $(\frac{-1}{p})=-1$ ?

(iv) if $y^2=x^3+7$ has an integral soln. (which it does not) then $x$ has to be odd and $y$ even, why is $y$ necessarily even?

(v) $y^2=x^3+7$ reducing equation modolu $4$ $\implies$ $0\equiv(x+3) \,[4]$ I see what they have done, e.g. $7\equiv3 \, [4]$ so 7 becomes 3.

how can we prove that $0\equiv(x+3) \, [4]$ is even true?

(vi) $x$ odd and $x+2>0$ and $x+2\equiv3 \, [4]$ $\implies$ it has prime factor $p\equiv3 \, [4]$ why?

thanks a bunch

Damien L
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Mathproof P.
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  • It is so much easier when you write $y^2 + 1 = x^3 + 8.$ See http://math.stackexchange.com/questions/330779/solve-y2-x3-33-in-integers/330806#330806 including my comment about $a^2 + b^2$ – Will Jagy Mar 16 '13 at 23:09
  • The explanation $\left(\frac{7}{8}\right)=-1$ is a strange way to put it. The author is using the Jacobi symbol, a generalization of the Legendre symbol. But it would be much easier to say that $7$ is of the form $4k+3$, so cannot be a perfect square. – André Nicolas Mar 16 '13 at 23:16
  • What Book? ${}{}{}$ – Will Jagy Mar 16 '13 at 23:33
  • As for iii), $\displaystyle \left(\frac{-1}{p}\right) = (-1)^{(p-1)/2} = 1 \iff p \equiv 1\ (4)$ for odd $p$. – J.H. Mar 16 '13 at 23:39

2 Answers2

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The following address your specific questions:

The explanation $\left(\frac{7}{8}\right)=-1$ is a strange way to put it. The author seems to be using a non-standard variant of the Jacobi symbol to indicate that a certain quadratic congruence is not solvable. It would have been much clearer to say that $7$ is of the form $4k+3$, so cannot be a perfect square.

A number certainly cannot be simultaneously congruent to $1$ and to $3$ modulo $4$: it cannot have simultaneously remainder $1$ and remainder $3$ on division by $4$. Any *odd number is congruent to exactly $1$ or $3$ modulo $4$.

For your next question, we cannot have $x$ even. For if $x$ is even then $x^3$ is divisible by $8$, so $x^3+7$ is congruent to $3$ modulo $4$, and therefore cannot be a square. Thus $x$ is odd. It follows that $x^3$ is odd, so $x^3+7$ is even. Thus $y^2$ is even, which forces $y$ to be even.

The notation now gets hard to read. But I think you are asking at the end why a positive integer which is congruent to $3$ modulo $4$ must have a prime factor which is congruent to $3$ modulo $4$. So suppose that $b\equiv 3\pmod{4}$, where $b$ is positive. We can assume that $b\gt 1$. Any product of two numbers congruent to $1$ modulo $4$ is itself congruent to $1$ modulo $4$. But $m$ can be expressed as a product of (not necessarily distinct) primes. If these were all congruent to $1$ modulo $4$, then their product $m$ would also be congruent to $1$ modulo $4$. But it isn't.

André Nicolas
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  • thanks for this, just small add on: ''A number certainly cannot be simultaneously congruent to 1 and to 3 modulo 4: it cannot have simultaneously remainder 1 and remainder 3 on division by 4'' if x cong. 1 and 3 mod 4 then this implies x is even (is does there exist such an x?), but the converse is false e.g. x=2. – Mathproof P. Mar 18 '13 at 13:31
  • sorry I get it. If x is cong. 1 and 3 mod 4 (this implies x is even) but x is odd by assumption. – Mathproof P. Mar 18 '13 at 13:39
  • In symbols, if we had $x\equiv 1\pmod{4}$ and $x\equiv 3\pmod{4}$, then we would have $3\equiv 1\pmod{4}$, so $2\equiv 0\pmod{4}$, which is false. If $a\equiv 0\pmod 4$ then $a\equiv 0\pmod{2}$, but the converse does not hold. I think that is what you are saying above, and it is true. Maybe thinking in terms of remainders is more concrete. – André Nicolas Mar 18 '13 at 13:51
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Here, I will try to be as explicit as possible to show that $y^2=x^3+7$ has no integer solution.

Case 1: Suppose that $x$ is even, so $x=2k$. It follows that $y^2=8k^3+7$. This implies that $y^2\equiv 7\mod 8$. This is a contradiction because there is no $y$ such that $y^2\equiv 7\mod 8$. You can check this by squaring the numbers $0,1,\ldots,7$, and showing that none of them give a remainder of $7$ when you divide by $8$.

Case 2: Suppose that $x$ is odd. Notice that $x^3+7$ will be even, so that $y$ must also be even (else $y^2=x^3+7$ will be odd). Note that $x^3\equiv x\mod 4$ (check this for $x=1,3$), $y^2\equiv 0\mod 4$ (because $y$ is even), and $7\equiv 3\mod 4$. Thus, reducing our equation mod $4$ gives the equation $0\equiv (x+3)\mod 4$, or $x\equiv 1\mod 4$.

Now write $y^2+1=x^3+8=(x+2)(x^2-2x+4)$. Since $x+2\equiv 3\mod 4$, there must exist some prime $p\equiv 3\mod 4$ dividing $x+2$. This follows from the fact that all odd numbers are congruent to either $1$ or $3\mod 4$, and the product of two numbers congruent to $1\mod 4$ is also congruent to $1\mod 4$. Then for this prime $p$ we can write $y^2+1\equiv 0\mod p$. This is impossible by quadratic reciprocity.

Hence, no integer solutions exist.

Jared
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