MHB Find the number of integers "k"

[anemone]

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

 

[mente oscura]

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)

There will be an easier way, not?

Spoiler

(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]

Hello.

Again.(Muscle)

There will be an easier way, not?

Spoiler

(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.

Spoiler

(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.

 

[anemone]

Hello.

Thank you very much.

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

Spoiler

(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.

 

[mente oscura]

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

Hello.

Excuse me.
Allow me to take part again. I have checked my calculations and have a mistake.

Spoiler

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.

 

[anemone]

Hello.

Excuse me.
Allow me to take part again. I have checked my calculations and have a mistake.

Spoiler

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.:)

 

[mente oscura]

Sorry, please try again.:)

Spoiler

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.:)

 

[Opalg]

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]

 

[anemone]

Spoiler

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.:)

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!

 

[mente oscura]

... 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.