# Find the number of integers "k"

#### anemone

##### MHB POTW Director
Staff member
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

##### Well-known member
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.

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

##### Well-known member
Hello.

Again.

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.

#### anemone

##### MHB POTW Director
Staff member
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.

#### mente oscura

##### Well-known member
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.
$$k=1253$$

The longest line:

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

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

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

$$k_1+k_2+k_3=1253$$

Regards.

#### anemone

##### MHB POTW Director
Staff member
Hello.

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

The longest line:

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

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

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

$$k_1+k_2+k_3=1253$$

Regards.

#### mente oscura

##### Well-known member
$$For : \ a=1 \ , \ b=0 \rightarrow{k_1=1006}$$

$$0 \le c \le 1005$$

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

It corresponds:

$$k: \ 1,3,5,...,2009,2011$$

$$For : \ a=2 \ , \ b=0 \rightarrow{k_2=250}$$

$$0 \le c \le 249$$

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

It corresponds:

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

Solution:

$$k_1+k_2=1256$$

Regards.

#### Opalg

##### MHB Oldtimer
Staff member

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 $1256.$

#### anemone

##### MHB POTW Director
Staff member
$$For : \ a=1 \ , \ b=0 \rightarrow{k_1=1006}$$

$$0 \le c \le 1005$$

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

It corresponds:

$$k: \ 1,3,5,...,2009,2011$$

$$For : \ a=2 \ , \ b=0 \rightarrow{k_2=250}$$

$$0 \le c \le 249$$

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

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.

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 $1256.$
Thank you Opalg for your well-explained solution!

#### mente oscura

##### Well-known member
... 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.