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The question is : Show that there are infinitely many pairs (a,b) of relatively prime integers (not necessarily positive ) such that both quadratic equations $x^2+ax+b$ =0 and $x^2+2ax+b$ =0 have integer roots. This is the problem .I tried it to do in many way. At first we let that there are finely many pairs then we shall show it is a contradiction. It is an INMO problem. Somebody please help me. Thank you

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    $\begingroup$ Hint: For integer $p,q$, $x^2+px+q=0$ has integer roots iff $p^2-4q$ is a square. $\endgroup$ – πr8 Sep 28 '16 at 16:08
  • $\begingroup$ @πr8 $$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ wouldn't always have integer roots if $b^2 - 4ac$ is a perfect square. $\endgroup$ – Parth Sep 28 '16 at 16:29
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    $\begingroup$ Yes it would if a=1 ( as in this case ) $\endgroup$ – user369582 Sep 28 '16 at 16:31
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    $\begingroup$ @unknowncoder given $a=1$ it always will $\endgroup$ – N.S.JOHN Sep 28 '16 at 16:31
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Hint: We just have to show that for an infinite number of integer couples $(a,b)$ with $\gcd(a,b)=1$ both $a^2-4b$ and $a^2-b$ are squares. If we set $$ a^2-b = (a-k)^2 = a^2-2ak+k^2 $$ we get $b=2ak-k^2$ and by imposing that $$ a^2-4b = a^2-8ak+4k^2 = (a-4k)^2 - 3 (2k)^2 $$ is a square, we get that the problem is equivalent to showing that there are an infinite number of rational points on the curve $\color{red}{x^2-3y^2=1}$. Since $(2,1)$ is a rational point, an infinitude of rational points is given by intersecting $y=m(x-2)+1$ (with $m\in\mathbb{Q}$) with $x^2-3y^2=1$.

So, long story short, this is just another exercise that can be tackled through Vieta jumping.

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  • $\begingroup$ Vieta jumping is overkill, though - there is a simple polynomial parameterized set of solutions. $\endgroup$ – Thomas Andrews Sep 29 '16 at 14:28
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Wanting $a^2-b=c^2$ and $a^2-4b=d^2$, we can see that we want $4c^2-d^2=3a^2$. If $a$ is odd, we can solve $2c-d=3$, $2c+d=a^2$ getting:

$$c=\frac{a^2+3}{4}, d=\frac{a^2-3}{2}$$

Then use $3b=c^2-d^2$ to determine $b$ in terms of $a$, and then do a lot of arithmetic to get an explicit infinite family. There is another (simple) condition required on $a$ to make $a,b$ relatively prime.

You get, after some work, that if $a=2k-3$ then $b=-k(k-1)(k-2)(k-3)$, then $$x^2+ax+b=0$$ has two integer solutions, $k(k-2)$ and $-(k-1)(k-3)$, and $$x^2+2ax+b=0$$ has two integer solutions $k^2-k$ and $-(k-2)(k-3)$.

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