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Find the number of integers "k"

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,678
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
Nov 29, 2013
172
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?

[tex](a^2+b^2+2c)(a-b)(a+b)=k[/tex]

Solution:

[tex]1 \le a \le 8[/tex]

[tex]0 \le b \le 7[/tex]

[tex]4 \le c \le 1005[/tex]

Supposing: a>b


Regards.

Edit. Misprint
 

mente oscura

Well-known member
Nov 29, 2013
172
Hello.

Again.(Muscle):mad:

There will be an easier way, not?

[tex](a^2+b^2+2c)(a-b)(a+b)=k[/tex]

Solution:

[tex]1 \le a \le 8[/tex]

[tex]0 \le b \le 7[/tex]

[tex]4 \le c \le 1005[/tex]

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.

[tex](a^2+b^2+2c)(a-b)(a+b)=k[/tex]

Solution:

[tex]1 \le a \le 8[/tex]

[tex]0 \le b \le 7[/tex]

[tex]0 \le c \le 1005[/tex]

[tex]k=1999[/tex]


Regards.

Edit.
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,678
Hello.

Thank you very much.

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

[tex](a^2+b^2+2c)(a-b)(a+b)=k[/tex]

Solution:

[tex]1 \le a \le 8[/tex]

[tex]0 \le b \le 7[/tex]

[tex]0 \le c \le 1005[/tex]

[tex]k=1999[/tex]


Regards.

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

mente oscura

Well-known member
Nov 29, 2013
172
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.
[tex]k=1253[/tex]

The longest line:

[tex]For : \ a=1 \ , \ b=0 \rightarrow{k_1=1006} [/tex]

[tex]For : \ a=2 \ , \ b=0 \rightarrow{k_2=245}[/tex]
Added:

[tex]For : \ a=3 \ , \ b=1 \rightarrow{k_3=2}[/tex]

[tex]k_1+k_2+k_3=1253[/tex]



Regards.
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,678
Hello.

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

The longest line:

[tex]For : \ a=1 \ , \ b=0 \rightarrow{k_1=1006} [/tex]

[tex]For : \ a=2 \ , \ b=0 \rightarrow{k_2=245}[/tex]
Added:

[tex]For : \ a=3 \ , \ b=1 \rightarrow{k_3=2}[/tex]

[tex]k_1+k_2+k_3=1253[/tex]



Regards.
Sorry, please try again.:)
 

mente oscura

Well-known member
Nov 29, 2013
172
Sorry, please try again.:)
[tex]For : \ a=1 \ , \ b=0 \rightarrow{k_1=1006} [/tex]

[tex]0 \le c \le 1005[/tex]

[tex]c: \ 0,1,2,...,1004,1005[/tex]

It corresponds:

[tex]k: \ 1,3,5,...,2009,2011[/tex]


[tex]For : \ a=2 \ , \ b=0 \rightarrow{k_2=250}[/tex]

[tex]0 \le c \le 249[/tex]

[tex]c: \ 0,1,2,...,248,249[/tex]

It corresponds:

[tex]k: \ 16,24,32,...,2000,2008[/tex]

Solution:

[tex]k_1+k_2=1256[/tex]


Regards.:)
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
I agree with mente oscura's latest version.

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 $1256.$
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,678
[tex]For : \ a=1 \ , \ b=0 \rightarrow{k_1=1006} [/tex]

[tex]0 \le c \le 1005[/tex]

[tex]c: \ 0,1,2,...,1004,1005[/tex]

It corresponds:

[tex]k: \ 1,3,5,...,2009,2011[/tex]


[tex]For : \ a=2 \ , \ b=0 \rightarrow{k_2=250}[/tex]

[tex]0 \le c \le 249[/tex]

[tex]c: \ 0,1,2,...,248,249[/tex]

It corresponds:

[tex]k: \ 16,24,32,...,2000,2008[/tex]

Solution:

[tex]k_1+k_2=1256[/tex]


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 mente oscura's latest version.

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
Nov 29, 2013
172
... 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.