Find the number of integers "k"

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Discussion Overview

The discussion centers around finding the number of integers \( k \) within the range \( 1 \le k \le 2012 \) that satisfy the equation \( a^2(a^2+2c)-b^2(b^2+2c)=k \) for non-negative integers \( a, b, c \). The scope includes mathematical reasoning and exploration of potential solutions.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants propose rewriting the equation as \( (a^4-b^4) + 2(a^2-b^2)c = k \) to analyze the problem.
  • One participant notes that for \( (a,b) = (1,0) \), the equation yields \( 1+2c = k \), generating all odd numbers from 1 to 2011, totaling 1006 numbers.
  • Another participant mentions that for \( (a,b) = (2,0) \), the equation gives \( 16+8c = k \), which generates all multiples of 8 except for 8 itself, totaling 250 numbers.
  • It is suggested that if both \( a \) and \( b \) are even or both odd, the results yield multiples of 8, while opposite parity results in odd numbers.
  • One participant calculates an overall total of 1256 valid integers \( k \) based on the above reasoning.
  • There are corrections and acknowledgments of earlier miscalculations by participants, indicating a collaborative effort to refine the understanding of the problem.

Areas of Agreement / Disagreement

Participants express differing views on the total count of integers \( k \) that can be generated, with some supporting the calculations leading to a total of 1256 while others have not reached a consensus on the correctness of the methods used.

Contextual Notes

Some participants acknowledge misprints and misunderstandings in earlier posts, indicating that the discussion is still evolving and that assumptions may not be fully clarified.

anemone
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Find the number of integers $k$ with $1 \le k \le 2012$ for which there exist non-negative integers $a, b, c$ satisfying the equation $a^2(a^2+2c)-b^2(b^2+2c)=k$.
 
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anemone said:
Find the number of integers $k$ with $1 \le k \le 2012$ for which there exist non-negative integers $a, b, c$ satisfying the equation $a^2(a^2+2c)-b^2(b^2+2c)=k$.

Hello.

Again.(Muscle):mad:

There will be an easier way, not?

(a^2+b^2+2c)(a-b)(a+b)=k

Solution:

1 \le a \le 8

0 \le b \le 7

4 \le c \le 1005

Supposing: a>b

Regards.

Edit. Misprint
 
mente oscura said:
Hello.

Again.(Muscle):mad:

There will be an easier way, not?

(a^2+b^2+2c)(a-b)(a+b)=k

Solution:

1 \le a \le 8

0 \le b \le 7

4 \le c \le 1005

Supposing: a>b

Regards.

Edit. Misprint

Hello.

Thank you very much.

Evidently, I did not understand well the question.
In addition, I am going to correct another misprint.

(a^2+b^2+2c)(a-b)(a+b)=k

Solution:

1 \le a \le 8

0 \le b \le 7

0 \le c \le 1005

k=1999

Regards.

Edit.
 
mente oscura said:
Hello.

Thank you very much.

Evidently, I did not understand well the question.
In addition, I am going to correct another misprint.

(a^2+b^2+2c)(a-b)(a+b)=k

Solution:

1 \le a \le 8

0 \le b \le 7

0 \le c \le 1005

k=1999

Regards.

Edit.

Thanks for participating mente oscura, but I'm sorry, your answer isn't correct.
 
anemone said:
Thanks for participating mente oscura, but I'm sorry, your answer isn't correct.
Hello.

Excuse me.:mad:
Allow me to take part again. I have checked my calculations and have a mistake.
k=1253

The longest line:

For : \ a=1 \ , \ b=0 \rightarrow{k_1=1006}

For : \ a=2 \ , \ b=0 \rightarrow{k_2=245}
Added:

For : \ a=3 \ , \ b=1 \rightarrow{k_3=2}

k_1+k_2+k_3=1253

Regards.
 
mente oscura said:
Hello.

Excuse me.:mad:
Allow me to take part again. I have checked my calculations and have a mistake.
k=1253

The longest line:

For : \ a=1 \ , \ b=0 \rightarrow{k_1=1006}

For : \ a=2 \ , \ b=0 \rightarrow{k_2=245}
Added:

For : \ a=3 \ , \ b=1 \rightarrow{k_3=2}

k_1+k_2+k_3=1253

Regards.

Sorry, please try again.:)
 
anemone said:
Sorry, please try again.:)

For : \ a=1 \ , \ b=0 \rightarrow{k_1=1006}

0 \le c \le 1005

c: \ 0,1,2,...,1004,1005

It corresponds:

k: \ 1,3,5,...,2009,2011For : \ a=2 \ , \ b=0 \rightarrow{k_2=250}

0 \le c \le 249

c: \ 0,1,2,...,248,249

It corresponds:

k: \ 16,24,32,...,2000,2008

Solution:

k_1+k_2=1256

Regards.:)
 
I agree with http://mathhelpboards.com/members/mente-oscura/'s latest version.

[sp]Write the equation as $(a^4-b^4) + 2(a^2-b^2)c = k$.

If $(a,b) = (1,0)$ then we get $1+2c = k$. That generates all the odd numbers, 1 to 2011, total $1006$ numbers.

If $(a,b) = (2,0)$ then we get $16+8c = k.$ That generates all the multiples of 8 except for 8 itself. Total $250$ numbers.

After that, there will be nothing new. If $a$ and $b$ are both even or both odd then we get multiples of 8. If they have opposite signs then we get odd numbers. So the overall total is $1256.$[/sp]
 
mente oscura said:
For : \ a=1 \ , \ b=0 \rightarrow{k_1=1006}

0 \le c \le 1005

c: \ 0,1,2,...,1004,1005

It corresponds:

k: \ 1,3,5,...,2009,2011For : \ a=2 \ , \ b=0 \rightarrow{k_2=250}

0 \le c \le 249

c: \ 0,1,2,...,248,249

It corresponds:

k: \ 16,24,32,...,2000,2008

Solution:

k_1+k_2=1256

Regards.:)

Yes, that is correct and well done! By the way, I hope you don't mind me asking you to keep plugging away at this problem.:)
Opalg said:
I agree with http://mathhelpboards.com/members/mente-oscura/'s latest version.

[sp]Write the equation as $(a^4-b^4) + 2(a^2-b^2)c = k$.

If $(a,b) = (1,0)$ then we get $1+2c = k$. That generates all the odd numbers, 1 to 2011, total $1006$ numbers.

If $(a,b) = (2,0)$ then we get $16+8c = k.$ That generates all the multiples of 8 except for 8 itself. Total $250$ numbers.

After that, there will be nothing new. If $a$ and $b$ are both even or both odd then we get multiples of 8. If they have opposite signs then we get odd numbers. So the overall total is $1256.$[/sp]

Thank you Opalg for your well-explained solution!
 
  • #10
anemone said:
... By the way, I hope you don't mind me asking you to keep plugging away at this problem.:)

Not. On the contrary. I am very grateful, of that one has offered more opportunities, to correct my "miscalculations". Thank you.:)

Regards.
 

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