How can I find a nontrivial perfect square of a given quadratic form?

Question

For example, say I wanted to find a perfect square of the form
z2 + 180z - 88

I happen to know through trial and error that a valid answer is z = 22, but how do I figure out this kind of problem in general?

Just to clarify, yes, the first term is raised to the second power, I sometimes omit the ^ symbol. Second, I'm not looking for a zero of the quadratic. I'm looking for an integer generated by this quadratic form that is in fact a perfect square.

Ah, and yes, the solution for z needs to be an integer.

Answer 1

To translate your problem, you want to find an integer x such that

z^2 + 180z - 88 = x^2

Let's complete the square on the left hand side.

z^2 + 180z + 8100 - 88 = x^2 + 8100
(z + 90)^2 - 88 = x^2 + 8100

Therefore, moving the x^2 to the left hand side, and the -88 to the right hand side,

(z + 90)^2 - x^2 = 8188

Factoring the left hand side as a difference of squares, we obtain

([z + 90] - x) ([z + 90] + x) = 8188

(z + 90 - x) (z + 90 + x) = 8188

Now, we determine the prime factorization of 8188.

8188 = 2 x 4094 = 2 x 2 x 2047 = 2 x 2 x 23 x 89

Note that both (z + 90 - x) and (z + 90 + x) must both be even. Out of the prime factors, the only two ways to do that is through the combination (2 x 23)(2 x 89) = (46)(176) and (2)(2x23x89) =
(2)(2 x 4094)

This gives us this only two possible combinations:

(z + 90 - x) (z + 90 + x) = (46)(176)
(z + 90 - x) (z + 90 + x) = (2)(4094)

Now, we can equate each of them componentwise and we get a system of equations.

(1) (z + 90 - x) (z + 90 + x) = (46)(176) means

z + 90 - x = 46
z + 90 + x = 176

(z = 21, x = 65)

(2) (z + 90 - x) (z + 90 + x) = (2)(4094) means

z + 90 - x = 2
z + 90 + x = 4094

(z = 1958, x = 2046)

Therefore, we get two values for z in order for z^2 + 180z - 88 to be a perfect square:

z = {21, 1958}

They create the corresponding perfect squares 65^2 and 2046^2.

(EDIT: Giving credit to "a_math_guy",
(z + 90 - x) (z + 90 + x) = 8188

implies that both (z + 90 - x) and (z + 90 + x) must be even.)

Answer 2

I'm going to assume you want z to be an integer, since if z can be anything, you can just solve z^2 + 180z - 88 = 1 (or 4 or 9 or etc.) using the quadratic formula.

One way to do it is to complete the square and try a factoring. You're looking for n,z that satisfy n^2 = z^2 + 180z - 88.
n^2 = (z + 90)^2 - 8188
Rearrange to get
(z + 90)^2 - n^2 = 8188
(z + 90 + n)(z + 90 - n) = 8188
At this point, factor 8188 and set one factor equal to z + 90 - n and the other factor equal to z + 90 + n. Then solve the simultaneous equations. A tip is to make the factors both odd or both even. If you make one factor even and the other odd, you'll get fraction solutions, which you don't want.

For example, 8188 = 2*4094
z + 90 + n = 4094
z + 90 - n = 2
2z + 180 = 4096
2z = 3916
z = 1958, n = 2046

You get other solutions depending on how you factor 8188. (To get the solution you found, use 8188 = 46*178.)

Answer 3

Here is a way, that I believe works for equations of the form x^2 + bx + c for all cases when b is even and (c - (b^2)/4) is either odd or divisible by 4 . First complete the square, that is, note that this equals:

(x + b/2)^2 + (c - (b^2)/4)

If (c - (b^2)/4) is odd and > 0, then call it 2n+1, then if you set x to be n - b/2, then the formula gives you (n+1)^2. Otherwise if (c - (b^2)/4) is odd and < 0, let x =n - b/2 + 1, and the formula gives n^2.

If however, (c - (b^2)/4) is divisible by 4 and >0, then call it 4n. Then set x to be n - b/2 - 1 and then the formula again gives you (n+1)^2. Otherwise, if (c - (b^2)/4) is divisible by 4 and < 0, let x be n - b/2 + 1, then the formula gives you (n-1)^2. For example, in the equation you gave: x^2 + 180x - 88, completing the square gives: (x + 90)^2 - 8188 = (x + 90)^2 - 4*2047. So if you let x = 2047 - 90 + 1 = 1958, then x^2 + 180x - 88 = 2046^2.

There must be a cleaner or more general way to do this, although it is not always possible. For example, there is no square of the form x^2 + 2.

Answer 4

Puggy was on the right track but should've added 88 to both sides so (z+90+x)(z+90-x)=8188 To make sure z and x are both integers, we want both factors z+90+x and z+90-x to be even. 8188=2*2*23*89 so set z+90+x=2*89 and z+90-x=2*89 to find z=22 and x (not that you care) = 66

For any rational number t, the value z=22*(88+t^2)/ (3872-132*t+ t^2) gives the value x=22*(-172*t- 264+3*t^2)/ (3872-132*t+ t^2).

Answer 5

22^2 + 180(22) - 88 ≠ 0
Something wrong?

------------
Daren,
"z^2+180z-88=(z^2+180z+(180/2)-...
=(z+90)^2-178 " is not right.

z^2 +180z - 88 = z^2 + 180z +90^2 - 90^2 - 88 = (z+90)^2 - 8188

Answer 6

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Answer 7

I suppose you mean z^2+180z-88.

z^2+180z-88
=(z^2+180z+(180/2)²-88-(180/2)²
=(z+90)^2-8188

(z+90)^2=8188
z+90=8188^1/2
z=0.487568206

sahsjing: refresh your browser,edited it already before you posted those..