Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given the following function: $$f(x) = \sqrt{36 x^2+16416 x +61084}$$ where x is an integer and the function also generates an integer value, is there an algorithm to determine its integer solutions?

share|improve this question

put on hold as off-topic by Daniel W. Farlow, Solid Snake, user91500, Claude Leibovici, Clayton Jul 29 at 16:52

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Daniel W. Farlow, user91500, Claude Leibovici, Clayton
If this question can be reworded to fit the rules in the help center, please edit the question.

    
What do you mean by "solution"? Do you mean the x for which $f(x) =0$? –  DavidButlerUofA Jul 16 '14 at 20:05
1  
Your question is a little unclear: are you asking for solutions to "$f(x) = 0$"? Or "$f(x)$ is an integer"? Or something else? –  Hurkyl Jul 16 '14 at 20:06
4  
The term $72*x*228$ also hints at a typo. –  hardmath Jul 16 '14 at 20:07
    
Sorry , I can not figure out what to do with the 228 –  bobbym Jul 16 '14 at 20:14
    
Sorry but the function f(x) generates an integer value. Also 72*x*228 = 16416*x. –  Nominal Jul 16 '14 at 21:00

2 Answers 2

up vote 1 down vote accepted

Given

$$ f(x) = \sqrt{ 36 x^2 + 16416 x + 61084}. $$

For what integer $x$ is $f(x)$ also an integer?


We can write

$$ f(x) = \sqrt{ \Big( 6 x + 1368 \Big)^2 - 1810340 }, $$

therefore

$$ \Big( 6 x + 1368 \Big)^2 - 1810340 = \Big( 6 x + 1368 - p \Big)^2, $$

so

$$ p \left\{ 2 \Big( 6 x + 1368 \Big) - p \right\} = 1810340. $$

Thus

$$ x = \left( \left( \frac{1810340}{p} + p \right) \Big/ 2 - 1368 \right) \Big/ 6. $$

Note the symmetry

$$ \frac{1810340}{p} + p = \frac{1810340}{1810340/p} + 1810340/p. $$

Using

$$ 1810340 = 2^2 \times 5 \times 7 \times 67 \times 193, $$

we can use limited values for $p$, like $p=2 \times 5 \times 7$, which is equivalent with $p= 2 \times 67 \times 193$.

We obtain the following results:

$$ \begin{array}{ccccc|c|c} 2 & 5 & 7 & 67 & 193 & p & x\\ \hline \times &&&&& 2 &75203\\ \times & \times &&&& 10 &14859\\ \times && \times &&& 14 &10549\\ \times & \times & \times &&& 70 &1933\\ \times &&& \times && 134 &909\\ \times &&&& \times & 386 &195\\ \times & \times && \times && 670 &53\\ \times & \times &&& \times & 1930 &11\\ \end{array} $$

share|improve this answer

You can write $f(x) = \sqrt{(6(x-228))^2 - 1810340}$.

Now, $f(x)$ is an integer if and only if $(6(x-228))^2 - 1810340$ is a square. Write $$k^2=(6(x-228))^2 - 1810340$$ so $$1810340 = (6(x-228))^2 - k^2 = (6x-6*228+k)(6x-6*228-k)$$

Now you factorize $1810340 = 2^2 * 5* 7* 67* 193$.

So you have to solve $48$ linear systems of the form $$ \left\{ \begin{matrix} 6x-6*228&+k &=& a \\ 6x-6*228&-k &=& b \end{matrix} \right. $$

For all possible values of $ab = 1810340$.

share|improve this answer
    
$f(6195)$ is in fact not an integer (I just plugged it into Wolfram). One potential issue I see with your proof is that $x$ and $k$ are not independent, so you cannot simply solve the last system of equations. –  Divergent Queries Jul 16 '14 at 21:18
    
No, I did a worse mistake. I made a mistake in completing the square. In fact, if you compute $(3*6195-228)^2-36713$ you get exactly $18356^2$. Ok, sorry, I made a mistake but I'm not erasing my answer. –  Crostul Jul 16 '14 at 21:37
    
That's fine. Mind if I edit it, though? –  Divergent Queries Jul 16 '14 at 21:39
    
I am editing right now. –  Crostul Jul 16 '14 at 21:41
1  
It should be $\sqrt{(6(x\color{red}{+}228))^2 - 1810340}$... –  johannesvalks Jul 16 '14 at 21:57

Not the answer you're looking for? Browse other questions tagged or ask your own question.