1

Question Statement:-

Show that the quadratic equation $x^2+7x-14(q^2+1)=0$, where $q$ is an integer, has no integral roots.

My attempt at a solution:-

For the given quadratic equation having rational coefficients to have integral roots the discriminant $D$ of the equation needs to be a perfect square.

$$\therefore D=(7)^2+56(q^2+1)=56q^2+105=7(8q^2+15)$$

Now, from the expression for the discriminant we see that $(8q^2+15)$ is always odd so for the discriminant to be a perfect square, the following condition needs to be satisfied

$$8q^2+15=7(2n+1)^2,\qquad\qquad\text{where n $\in Z$}$$

So $$2(q^2+1)=7(n^2+n)$$

After this I was having trouble coming up with a way to disprove the above equality that I came up with, if you could help me out as to how to proceed from here on or maybe suggest a way that makes it easier for me to prove what the question asks.

Community
  • 1
user350331
  • 2,728
  • 4
  • 20
  • 41

5 Answers5

5

If $7(8q^2+15)$ is a perfect square, $8q^2+15$ is divisible by $7$, which is equivalent to $q^2 \equiv -1 \pmod 7$. This is well known to be impossible.

MooS
  • 31,390
1

If $x$ is a root of $x^2+7x-14(q^2+1)=0$, then $x$ is a multiple of $7$ because $x^2=7(-x+2(q^2+1))$.

Write $x=7y$. Then $7(y^2+y)=2(q^2+1)$ and so $7$ divides $q^2+1$. But this cannot happen because $7 \equiv 3 \bmod 4$. Or you can just check all possible residues of $q \bmod 7$ and see that $q^2+1$ is never $0$.

lhf
  • 216,483
0

$$8q^2+15=7m^2\Rightarrow q^2+1\equiv 0\pmod 7$$ but $q^2=-1$ has no solution in $\Bbb F_7$ because $(\Bbb F_7)^2=\{1,2,4,0\}$.

Actually $8q^2+15$ never is a multiple of $7$.

Piquito
  • 29,594
  • I do see that all the solutions contain the use of modulo but yours uses something more, which blew my mind even further away ;P. Can you please explain just the modulo part in detail. I dont know much about modulo except that it calculates remainder. – user350331 Nov 11 '16 at 12:59
  • Can you tell me, why is that, that $8q^2+15$ is never a multiple of 7? – user350331 Nov 11 '16 at 13:02
  • With pleasure. $8q^2+15=7(q^2+2)+q^2+1$ therefore the remainder you mention modulo $7$ is $q^2+1$ and $7m^2$ has remainder $0$. Besides the squares modulo $7$ of $1,2,3,4,5,6,0$ are respectively equal to $1,4,2,2,4,1,0$ so any square is equal to $6=-1$. – Piquito Nov 11 '16 at 13:07
  • @user350331: because my deduction does not pay attention to the square of $m$, it is just because RHS is a multiple of $7$. – Piquito Nov 11 '16 at 13:10
  • As $q^2+1=7(m^2-q^2-2) \implies q^2+1\equiv 0\pmod 7$. So, what this basically means is that either $q^2=6$ or $-1$, but both are not possible. Am I right, just confirming what I got from your solution, because I was not able to understand halfway what you tried to tell me in your comment. – user350331 Nov 11 '16 at 13:18
  • $6=-1$ is not an usual equality of integers but an equality in $\Bbb F_7$ i.e. an equality of classes modulo $7$, you have $6\equiv -1\pmod 7\iff 6+1=7\equiv 0\pmod 7$. Good by. No more comments. – Piquito Nov 11 '16 at 13:28
0

If $x^2+7x-14(q^2+1)=0$, where $x,q\in\mathbb Z$, then $$7\mid 7\left(2\left(q^2+1\right)-x\right)=x^2$$ so by Euclid's Lemma $7\mid x$, so $7^2\mid x^2$ and $7^2\mid 7x$, so $$7^2\mid x^2+7x=7\left(2\left(q^2+1\right)\right),$$

so $7\mid 2\left(q^2+1\right)$. Since $\gcd(7,2)=1$, by Euclid's Lemma $7\mid q^2+1$, i.e. $q^2\equiv -1\pmod{7}$, which is impossible, because $0^2\equiv 0$, $(\pm 1)^2\equiv 1$, $(\pm 2)^2\equiv 4$, $(\pm 3)\equiv 2$ mod $7$, so $-1$ isn't a quadratic residue mod $7$.

Another way here is noticing that $7$ is a prime of the form $4k+3$, $k\in\mathbb Z$, so by Quadratic Reciprocity (or a separate lemma, which is a lot easier to prove than quadratic reciprocity) $-1$ isn't a quadratic residue mod $7$.

user236182
  • 13,324
0

Since congruence arithmetic is unfamiliar we give a solution avoiding them.

Note $\ 7\mid 2m\,\Rightarrow\, 7\mid 7m-3(2m) = m.\,$ So $\,\ 7\mid 2(1+q^2)\,\Rightarrow\, 7\mid 1+q^2$

Dividing $\,q\,$ by $\,7\,$ we have $\, q = a + 7b\,$ with remainder $\,a\,$ satisfying $\,0\le a \le 6$

Thus $\ 1+q^2 = 1+a^2+14ab+49b^2 = 1+a^2 + 7(2ab+7b^2)$

so $\ 7\mid 1+q^2\,\Rightarrow\, 7\mid 1+a^2\,$ for some $a$ with $\,0\le a\le 6,\,$ a contradiction (check all $7$ cases).

Bill Dubuque
  • 272,048