Algebra: Find all integer values of x, such that sqrt(x^2+x+1) is rational

[Stas]

Hi! I've got an exercise and i couldn't solve it about 2 hours.

The exercise is : Find all integer values of x, such that sqrt(x^2+x+1) is rational.

I tried some approaches as assuming that x^2+x+1 can be represented as k^2, where k is rational, but after some algebra, i did not succeed.

 

[tkhunny]

Hi! I've got an exercise and i couldn't solve it about 2 hours.

The exercise is : Find all integer values of x, such that sqrt(x^2+x+1) is rational.

I tried some approaches as assuming that x^2+x+1 can be represented as k^2, where k is rational, but after some algebra, i did not succeed.

Great. Let's see some of that algebra.

\(\displaystyle If\;\sqrt{x^{2} + x + 1}\;is\;rational,\;what\;does\;that\;say\;of\;x^{2} + x+ 1?\;\;It\;can\;be\;represented\;as\;k^{2}.\) Excellent!

Now, can you solve the quadratic equation suggested by this revelation?

 

[Dr.Peterson]

Hi! I've got an exercise and i couldn't solve it about 2 hours.

The exercise is : Find all integer values of x, such that sqrt(x^2+x+1) is rational.

I tried some approaches as assuming that x^2+x+1 can be represented as k^2, where k is rational, but after some algebra, i did not succeed.

This is a nice problem, and tkhunny's suggestion is a very good one. Don't miss the fact that since x^2+x+1 will be an integer, your k will not only be rational, but an integer (since the square root of an integer is always either an integer or irrational).

If you are still having trouble, it may be a good idea to tell us the context of the problem, which is more than just an ordinary algebra problem. What have you learned recently that might be of use? I'm guessing it might be either the quadratic formula, or the discriminant.

 

[Stas]

This is a nice problem, and tkhunny's suggestion is a very good one. Don't miss the fact that since x^2+x+1 will be an integer, your k will not only be rational, but an integer (since the square root of an integer is always either an integer or irrational).

If you are still having trouble, it may be a good idea to tell us the context of the problem, which is more than just an ordinary algebra problem. What have you learned recently that might be of use? I'm guessing it might be either the quadratic formula, or the discriminant.

I've got a new math teacher and she gave us a new material about how to prove an expreesion is irational.
At this step i already know about quadratic formula and discriminant.

Back to this problem. I've got a nice idea as assuming that x^2+x+1 is a perfect square, so it can be expressed as (x+k)^2. The remaining term shoul be equal to 0, so we get that x can be expressed as (k^2-1)/(1-2k) where k should be an integer. So i trasnformed the main problem in a problem of type, find such a integer k, so the expression is integer, but at this step i couldnt proceed.

Back to the main method, today i got a knew ideea. So after writing x^2+x+1 as k^2, where k is integer, we can write this as
x^2+x=k^2-1 or x(x+1)=(k-1)(k+1).

There we analyze some variants:
1)If x=k-1, then x+1=k+1 or x=k, so there is no solutions
2)If x=k+1 then x+1=k-1 or x=k-2, so there is no solutions
3)If x=1, then x+1=(k-1)(k+1) or 2=k^2-1 or k^2=3, but we know that k is integer so also no solutions
4)If x+1=1, then x=0, so x=(k-1)(k+1) or k^2=1 or k=+-1, so x=0 is a valid solution
The last variant is if one of the parts is equal to 0, we get
x(x+1)=0, there can be 2 cases, if x=0 (this one we verified in the 4 point) and the second case if x+1=0, the x=-1 and k=+-1 so this is also a valid solution.

So i got the answer -1 and 0, that looks like correct,
but at this step i dont know if this is a good solution or there are no more variants.

 

[Dr.Peterson]

I've got a new math teacher and she gave us a new material about how to prove an expreesion is irational.
At this step i already know about quadratic formula and discriminant.

Back to this problem. I've got a nice idea as assuming that x^2+x+1 is a perfect square, so it can be expressed as (x+k)^2. The remaining term shoul be equal to 0, so we get that x can be expressed as (k^2-1)/(1-2k) where k should be an integer. So i trasnformed the main problem in a problem of type, find such a integer k, so the expression is integer, but at this step i couldnt proceed.

Back to the main method, today i got a knew ideea. So after writing x^2+x+1 as k^2, where k is integer, we can write this as
x^2+x=k^2-1 or x(x+1)=(k-1)(k+1).

There we analyze some variants:
1)If x=k-1, then x+1=k+1 or x=k, so there is no solutions
2)If x=k+1 then x+1=k-1 or x=k-2, so there is no solutions
3)If x=1, then x+1=(k-1)(k+1) or 2=k^2-1 or k^2=3, but we know that k is integer so also no solutions
4)If x+1=1, then x=0, so x=(k-1)(k+1) or k^2=1 or k=+-1, so x=0 is a valid solution
The last variant is if one of the parts is equal to 0, we get
x(x+1)=0, there can be 2 cases, if x=0 (this one we verified in the 4 point) and the second case if x+1=0, the x=-1 and k=+-1 so this is also a valid solution.

So i got the answer -1 and 0, that looks like correct,
but at this step i dont know if this is a good solution or there are no more variants.

Your answer is correct, but I think the discriminant makes it easier.

We have x^2 + x + 1 = k^2, where x and k are both integers. Solving x^2 + x + (1 - k^2) = 0, the discriminant is 1 - 4(1 - k^2) = 4k^2 - 3. For x to be rational, this must be a perfect square; so 4k^2 - 3 = n^2 for some integer n. Continue from here by factoring. (Maybe this isn't much easier than yours; but I had no concerns about missing anything when I did this.)

There may be a better way!

 

[JeffM]

Your answer is correct, but I think the discriminant makes it easier.

We have x^2 + x + 1 = k^2, where x and k are both integers. Solving x^2 + x + (1 - k^2) = 0, the discriminant is 1 - 4(1 - k^2) = 4k^2 - 3. For x to be rational, this must be a perfect square; so 4k^2 - 3 = n^2 for some integer n. Continue from here by factoring. (Maybe this isn't much easier than yours; but I had no concerns about missing anything when I did this.)

There may be a better way!

The discriminant plus factoring makes for a very simple solution given the knowledge that we want only integer answers.

\(\displaystyle 4k^2 - 3 = n^2 \implies 4k^2 - n^2 = 3 \implies (2k - n)(2k + n) = 3.\)

\(\displaystyle \therefore max(|2k + n|, |2k - n| \le 3 \ \because k,\ n \in \mathbb Z.\)

\(\displaystyle \therefore |2k| \le 3 \implies |k| < 2 \implies k = 1,\ 0, \text { or } -\ 1.\)

\(\displaystyle 4 * 0^2 - 3 < 0 \implies x \not \in \mathbb R \implies k \ne 0.\)

\(\displaystyle \therefore k = \pm 1 \implies k^2 = 1 \implies \sqrt{4k^2 - 3} = 1.\)

\(\displaystyle \therefore x = \dfrac{-\ 1 \pm 1}{2} \implies x = \dfrac{0}{2} \text { or } \dfrac{-\ 2}{2} \implies x = 0 \text { or } -\ 1.\)

 

[Stas]

The discriminant plus factoring makes for a very simple solution given the knowledge that we want only integer answers.

\(\displaystyle 4k^2 - 3 = n^2 \implies 4k^2 - n^2 = 3 \implies (2k - n)(2k + n) = 3.\)

\(\displaystyle \therefore max(|2k + n|, |2k - n| \le 3 \ \because k,\ n \in \mathbb Z.\)

\(\displaystyle \therefore |2k| \le 3 \implies |k| < 2 \implies k = 1,\ 0, \text { or } -\ 1.\)

\(\displaystyle 4 * 0^2 - 3 < 0 \implies x \not \in \mathbb R \implies k \ne 0.\)

\(\displaystyle \therefore k = \pm 1 \implies k^2 = 1 \implies \sqrt{4k^2 - 3} = 1.\)

\(\displaystyle \therefore x = \dfrac{-\ 1 \pm 1}{2} \implies x = \dfrac{0}{2} \text { or } \dfrac{-\ 2}{2} \implies x = 0 \text { or } -\ 1.\)

Wow this is so much clever, thanks a lot for a good explanation!

 

[Dr.Peterson]

The discriminant plus factoring makes for a very simple solution given the knowledge that we want only integer answers.

\(\displaystyle 4k^2 - 3 = n^2 \implies 4k^2 - n^2 = 3 \implies (2k - n)(2k + n) = 3.\)

\(\displaystyle \therefore max(|2k + n|, |2k - n| \le 3 \ \because k,\ n \in \mathbb Z.\)

\(\displaystyle \therefore |2k| \le 3 \implies |k| < 2 \implies k = 1,\ 0, \text { or } -\ 1.\)

\(\displaystyle 4 * 0^2 - 3 < 0 \implies x \not \in \mathbb R \implies k \ne 0.\)

\(\displaystyle \therefore k = \pm 1 \implies k^2 = 1 \implies \sqrt{4k^2 - 3} = 1.\)

\(\displaystyle \therefore x = \dfrac{-\ 1 \pm 1}{2} \implies x = \dfrac{0}{2} \text { or } \dfrac{-\ 2}{2} \implies x = 0 \text { or } -\ 1.\)

The approach I had in mind was a little different in the details. Since we know that n and k are integers, and can assume that n is positive, and since 3 factors only as 1*3 or -1*-3, from \(\displaystyle (2k - n)(2k + n) = 3\) we can see that \(\displaystyle 2k - n = 1 \text{ and } 2k + n = 3\), or \(\displaystyle 2k - n = -3 \text{ and } 2k + n = -1\). Therefore k = 1, or k = -1. Then x = 0 or -1.