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Consider a quadratic equation;

$$ x^2 + 7x – 14(a^2 + 1) = 0,$$

… (where $a$ is an integer)

For how many different value of $a$, the equation will have at least one integer root?

I found out its discriminant, it comes out to be

$$ (49 + 56(a^2+1))^{1/2}. $$ This should be the perfect square and also odd so that the at least one root be integer.

But I am unable to get the values.

How I can achieve this ?

Thanks in advance.

MJD
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vikiiii
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3 Answers3

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So, the discriminant $(D^2)$ is $49+56(a^2+1)=7(8a^2+15)$

As $7\mid D^2\implies 7\mid D$ if $D$ is an integer,

in that case, $49\mid D^2$ i.e., $49\mid \{49+56(a^2+1)\}$

$\implies 7\mid (8a^2+15)\implies 7\mid (a^2+1)$

$-1$ needs to be a quadratic residue of $7$.

Using Euler's Criterion,
$-1$ is a quadratic residue of primes $\equiv 1\pmod 4$, but $7\equiv -1\pmod 4$,

Without using quadratic residue, all the integers can be written as $7b,7b\pm1,7b\pm2,7b\pm 3$

So, $(7b)^2\equiv 0\pmod 7,(7b\pm1)^2\equiv 1\equiv -6,$ $(7b\pm 2)^2\equiv 4\equiv -3, (7b\pm3)^2\equiv 2\equiv -5$, so there is no solution to $a^2\equiv -1\pmod 7$

So, there is no rational solution of the given equation for the integral values of $a$.

lab bhattacharjee
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By Eisenstein's criterion we need $a^2+1$ to be divisible by 7 for the equation to have a rational root. The squares mod 7 are 0,1,2,4, and do not include -1, so this cannot happen.

Mark Bennet
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  • Can you explain me Eisenstein's Criterion for this question. – vikiiii Oct 20 '12 at 17:35
  • For this question André Nicholas and lab bhattacharjee have shown how Eisenstein cashes out without mentioning it. For a quadratic, if the coefficient of the highest term $x^2$ is not divisible by $p$ (7 in this case, and the coefficient is 1), and the coefficients of the other terms are divisible by $p$ (term in $x$ is divisible by 7, as is constant term) then there can only be a rational solution (integer here, in fact) if the constant term is divisible by $p^2$ ie 49. The others have shown that this must be the case. – Mark Bennet Oct 20 '12 at 17:45
  • To extend a little - Eisenstein is not necessary for this problem, but it is one of those things it is useful to know about as work becomes more advanced. – Mark Bennet Oct 20 '12 at 17:48
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If there is an integer solution, both (conceivably equal) solutions are integers, since their sum is $-7$. Moreover, since $7$ divides the last two terms, it must divide any integer solution $x$.

So let the solutions be $7a$ and $7b$. The product of the solutions is $49pq$. But it is also $-14(a^2+1)$. Now use Mark Bennet's calculation that shows that $a^2+1$ cannot be divisible by $7$.

André Nicolas
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