Find all integers x for which x^2 + 7x + 1 is a perfect square.?

Question

Please prove that your list is exhaustive.

Answer 1

oh i love these kind of questions!
so,we want to derive integer solutions for following equation:

x^2 + 7x + 1 = k^2

x^2 + 7x + (1 - k^2) = 0

if we solve this 2nd degree equation for x,we have its Discriminant as below:

49 - 4(1 - k^2)

so,if we want Integer solution,the discriminant must be a perfect square say y^2 (because if not,the 2nd root of the discriminant would be an irrational number and then x would be irrational too...) :

49 - 4(1 - k^2) = y^2
49 - 4 + 4k^2 = y^2
=>
y^2 - 4k^2 = 45
=>
(y - 2k) (y + 2k) = 45
since the divisors of 45 is finite,so the solutions would be finite too:
since (y + 2k) > (y - 2k) the only possible cases are when:
1)
y - 2k = 1 and y + 2k = 45

=> y = 23 and k = 11
=> x^2 + 7x + 1 = 121
=> x = 8
1st solution = ( 8 , 11)
=> x = -15
2nd solution = ( -15 , 11)

2)
y - 2k = 3 and y + 2k = 15

=> y = 9 and k = 3
=> x^2 + 7x + 1 = 9
=> x = 1
3rd solution = (1 , 3)
=> x = -8
4th solution = ( -8 , 3)

3)
y - 2k = 5 and y + 2k = 9

=> y = 7 and k = 1
=> x^2 + 7x + 1 = 1
=> x = 0
5th solution = ( 0 , 1 )
=> x = -7
and the last solution = ( -7 , 1)

so we got all solutions: (8 , 11) , ( -15 , 11) , (1 , 3) , ( -8 , 3)
, (0 , 1) , ( -7 , 1)

this is a good method to obtain integer solutions for some 2nd Degree Diophantine equations...

hope it helped...good luck!

Answer 2

Hi,

x^2 + 7x + 1 is a perfect square for x values of -15, -8, -7, 0, 1, and 8. These values give square roots of 1, 3, and 11. I do not see a pattern here and do not see how to prove that this list is exhaustive. I hope someone can show us how.

I hope that helps!! :-)

My compliments to Farbod. That answer is excellent!!!

Answer 3

It is not a perfect square

Completing the square

x² + 7x + 1 = 0

x² 7x + 1 - 1 = 0 - 1

x² + 7x = - 1

x² + 7x + ______= - 1 + ______

x² + 7x + 49/4 = - 1 + 49/4

(x + 7/2)(x + 7/2) = - 4/4 + 49/4

(x + 7/2)² = 45/4

(√x + 7/2) = ± √45 / √4

x + 7/2 = ± √45 / 2

x + 7/2 - 7/2 = - 7/2 ± √45 /2

x = - 7/2 ± √45 / 2

x = - 7/2 ± 0.6708203933 / 2

- - - - - - - - -

Solving for +

x = - 7/2 + 0.6708203933 / 2

x = - 0.291796067 / 2

x = - 0.145898034

- - - - - - - - - -

solving for -

x = - 7/2 - 6.708203933 / 2

x = - 13.70820393 / 2

x = - 6.8541001967

- - - - - - -

This problem can be solved by using the quadratic formula. The roots will be the same.

- - - - - - s-

Answer 4

You have x^2+7x+1=y^2. So 7x+1=(y-x)(y+x).
Since y>x, y+x > 2x and 7x+1less than 8x. So y-x<4. We can try y-x =1,2,or 3. If (y-x)=1, you get 7x+1=2x+1so x=0. If y-x=2, you get 2(2x+2)=7x+1 so x=1 and for y-x=3, you get 7x+1= 3(2x+3). So x=8. That's it

0,1,8.

Answer 5

1
I think it would be it