Dividing quadratic with linear

[rajeshmarndi]

I have two table, first one changes quadratically and second table changes linearly.

46 ... 55
90 ... 57
142 ... 59
202 ... 61
270 ... 63
346 ... 65
430 ... 67
522 ... 69
622 ... 71
730 ... 73
846 ... 75
970 ... 77

and so on.

First table data is quadratic and it is 4x^2 + 32x + 10 and
second table is linear and it is 2x + 53

I want to know which data in second table divide the first table data exactly (i.e give a whole number).

I know it is 73(10th row data) that divide 730 by 10. But is there any way I can know mathematically.

When I divide 4x^2 + 32x + 10 by 2x + 53
I get answer as 2x - 37 and remainder 1971(constant). But it give me no information that when x=10, 73 divide 730 exactly.

Thank you.

 

[symbolipoint]

The aim you describe is unsure. Look for the quadratic relationship between the two columns of data. Using graph paper to plot the points should help to identify a good quadratic fit. Use y for the first column values and use x from the second column corresponding values. Maybe try to EXPECT the quadratic relationship, and plot square roots of y on the vertical axis and the x values for the horizontal axis (column 2). This should look like a LINE, if the fit is truly quadratic between x and y.

 

[mfb]

@symbolipoint: OP has the relation already, and looks for rows where the first value is an integer multiple of the second.

I get answer as 2x - 37 and remainder 1971(constant)

Then you have to figure out when 1971 is a multiple of 2x + 53.

1971 = 3³*73. That allows to find all factors, all larger than 53 will lead to a solution.

 

[micromass]

So you are looking for integers $n,m$ such that

$$4n^2 + 32n + 10 = 2nm + 53m$$

Note that $m$ is always even, so we can write $m= 2M$. Our equation can then be written as:

$$(2n + 8-M)^2 - 54 -M^2 - 90M = 0$$

We set $a = 2n + 8 - M$, then we get

$$a^2 = (M + 45)^2 - 1971$$

We set $b = M + 45$. We get:

$$a^2 - b^2 = - 1971$$

or

$$(a-b)(a+b) = - 1971$$

So letting $d_1 = a-b$ and $d_2 = a+b$, we get $d_1d_2 = -1971$.

Now, $-1971$ has positive divisors $1,~3,~9,~27,~73,~219,~657,~1917$. So we have the following products that form $-1971$:

$$\begin{array}{|cc|cc|c|cc|}<br /> \hline<br /> d_1 & d_2 & a & b & M & n & m\\<br /> \hline<br /> -1 & 1971 & 985 & 986 & 941 & 959 & 1882\\<br /> 1 & -1971 & -985 & -986 & -1031 & -1012 & -2062\\<br /> -3 & 657 & 327 & 330 & 285 & 302 & 570\\<br /> 3 & -657 & -327 & -330 & -375 & -355 & -750\\<br /> -9 & 219 & 105 & 114 & 69 & 83 & 138\\<br /> 9 & -219 & -105 & -114 & -159 & -136 & -138\\<br /> -27 & 73 & 23 & 50 & 5 & 10 & 10\\<br /> 27 & - 73 & -23 & -50 & -95 & -63 & -190\\<br /> -73 & 27 & -23 & 50 & 5 & -13 & 10\\<br /> 73 & -27 & 23 & -50 & -95 & -40 & -190\\<br /> -219 & 9 & -105 & 114 & 69 & -22 & 138\\<br /> 219 &-9 & 105 & -114 & -159 & -31 & -318\\<br /> -657 & 3 & -327 & 330 & 285 & -25 & 570\\<br /> 657 & -3 & 327 & -330 & -375 & -28 & - 750\\<br /> -1971 & 1 & -985 & 986 & 941 & -26 & 1882\\<br /> 1971 & -1 & 985 & -986 & -1031 & -27 & -2062\\<br /> \hline<br /> \end{array}$$

If you care only about the positive $n$, then the only solutions are $959,~302,~83,~10$. Indeed:

$$\begin{array}{|c|cc|c|}<br /> \hline<br /> n & 2n+53 & 4n^2 + 32n + 10 & \text{quotient}\\<br /> \hline<br /> 959 & 1971 & 3709422 & 1882\\<br /> 302 & 657 &374490&570\\<br /> 83 & 219 & 30222&138\\<br /> 10 & 73 & 730&10\\<br /> \hline<br /> \end{array}$$