the question
Determine all the integers $x$ that have the property that $\sqrt{x^2+7x+21}$ is a rational number.
the idea
The number would be rational only if $x^2+7x+21$ would be a square number which means that $x^2+7x+21=k^2 => k^2-x^2=(x+k)(k-x)=7(x+3)$.
From here i thought of going with divisibilities, but got nothing useful. Idk what to do froward. Hope one of tou can help me! Thank you!
On
Just use that there cannot be a perfect square between two consecutive perfect squares (otherwise they would not be consecutive...) Then, for your polynomial $p(x)=x^2+7x+21$ it is easy to see that for any integer $x>5$ you will have: $$ (x+3)^2 < p(x) < (x+4)^2, $$ so in that case p(x) itself cannot be a perfect square. With some minus-signs or other trickery you'll find a similar rule for negative $x$-values below a certain negative number (I leave the details to you!) That leaves just a small region of possible candidates around 0 to check by hand.
NB: by consecutive we mean next neigbors in the sequence of perfect squares, not that they have to be true neigbors of course (that happens only for 0 and 1!)
PS: I see @Aig beat me on this!
On
For some integer $y$, we have $(2y)^2=4(x^2+7x+21)=(2x+7)^2+35$. Factorizing the difference of squares then gives four possible values for $\{2y+2x+7,2y-2x-7\}$: namely $\{\pm35,\pm1\}$ and $\{\pm7,\pm5\}$. Subtraction to eliminate $y$ correspondingly yields four values for $4x+14$, namely $\pm34$ and $\pm2$. Hence the possible values for $x$ are $-12,-4,-3,$ and $5$.
While there is a solution by completing a square, I want to present another way. We have that $x^2+7x+21$ is a square of an integer.
Let $x>0$. Then $$(x+3)^2< x^2+7x+21< (x+5)^2.$$ It means that $$x^2+7x+21=(x+4)^2.$$ This gives $x= 5$. Now let $x\le-9$. Then $$(x+4)^2< x^2+7x+21< (x+3)^2+12<(x+2)^2.$$
So we have that $$x^2+7x+21=(x+3)^2.$$ This gives $x= -12$. Now we just try all $x$ from $\{-8,-7,-6,-5,-4,-3,-2,-1,0\}$ and find that $x=-3$ and $x=-4$ work, too. So the answer is $x\in\{-12,-4,-3,5\}$.