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Question:

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

My approach :

Let for any integer $x$ the quadratic equation $=0$, Then $x(x-7)=14(q^2+1)$. We easily observe $2\mid x(x-7)$ . Now $7\mid x$ or $7\mid {(x-7)}$. In any of the cases $x=7m$ ($m$ belongs to integer) . Thus $7m(m-1)=2(q^2+1)$ , so $7 \mid {(q^2+1)}$ , therefore $q^2+1=7t$ or $q^2=7n+6$ Now if somehow we can prove that no perfect square could be represented as $7n+6$, By contradiction we will prove the question .

My problem

  • how do we prove that no perfect square could be represented as $7n+6$.

  • Please correct if my proposed solution is wrong at any step.

  • Other elegant solutions will also be highly appreciated.

Thanks..!

Akshaj Bansal
  • 970

2 Answers2

1

To prove that no perfect square $k^2=7n+6$, we work with $k=7m\pm l; l\in0,1,2,3$

$$(7m+0)^2=49m^2=7(7m^2)+0$$ $$(7m\pm1)^2=49m^2\pm14m+1=7(7m^2\pm2m)+1$$ $$(7m\pm2)^2=49m^2\pm28m+4=7(7m^2\pm4m)+4$$ $$(7m\pm3)^2=49m^2\pm42m+9=7(7m^2\pm6m+1)+2$$

None of these are of the form $7n+6$, thusly it is not possible

Rhys Hughes
  • 12,842
0

$$x^2-7x-14(q^2+1)=0 \implies \frac{7\pm \sqrt{49+56(q^1+1)}}{2}$$ In oder the roots are integer they need to be rational first. The discriminant $105+56q^2$ needs to be perfect square of an integer ($n$). Then $$56q^2-n^2=105$$ $q$ and $n$ being integers RHS is can have rational factors but LHS does not have rational factors like $a^2-b^2=(a-b)(a+b)$. So the roots will be real but they cannt be even rational, so we conclude no integer roots.

Z Ahmed
  • 43,235