MHB Counting Integers k with Satisfying Equations Involving Non-Negative Integers

[anemone]

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

($a,\,b,\,c$ are not necessarily distinct.)

 

[Opalg]

[sp]$$k = a^2(a^2+2c)-b^2(b^2+2c) = a^4 - b^4 + 2c(a^2 - b^2) = (a^2 - b^2)(a^2+b^2+2c).$$ If $a=1$ and $b=0$ then $k = 1+2c$. In that way, $k$ can be any odd number apart from $1$.

If $a=2$ and $b=0$ then $k = 8(2+c)$. In that way, $k$ can be any multiple of $8$ apart from $8$ itself.

To see that these are the only values of $k$ that can occur, notice that if $a$ and $b$ have opposite signs then $a^2-b^2$ and $a^2+b^2+2c$ will both be odd, and therefore $k$ will be odd. If $a$ and $b$ are both even, or both odd, then $a^2-b^2$ will be a multiple of $4$ and $a^2+b^2+2c$ will be even, and so $k$ will be a multiple of $8$.

There are $1005$ odd numbers between $2$ and $2011$ (inclusive), and there are $250$ multiples of $8$ between $16$ and $2008$ (inclusive), giving a total of $1255$ possible values for $k$.[/sp]

 

[anemone]

[sp]$$k = a^2(a^2+2c)-b^2(b^2+2c) = a^4 - b^4 + 2c(a^2 - b^2) = (a^2 - b^2)(a^2+b^2+2c).$$ If $a=1$ and $b=0$ then $k = 1+2c$. In that way, $k$ can be any odd number apart from $1$.

If $a=2$ and $b=0$ then $k = 8(2+c)$. In that way, $k$ can be any multiple of $8$ apart from $8$ itself.

To see that these are the only values of $k$ that can occur, notice that if $a$ and $b$ have opposite signs then $a^2-b^2$ and $a^2+b^2+2c$ will both be odd, and therefore $k$ will be odd. If $a$ and $b$ are both even, or both odd, then $a^2-b^2$ will be a multiple of $4$ and $a^2+b^2+2c$ will be even, and so $k$ will be a multiple of $8$.

There are $1005$ odd numbers between $2$ and $2011$ (inclusive), and there are $250$ multiples of $8$ between $16$ and $2008$ (inclusive), giving a total of $1255$ possible values for $k$.[/sp]

Well done Opalg!(Yes)(Yes) And thanks for participating!:)