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- Feb 14, 2012

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Find all solutions to $(m^2+k)(m+k^2)=(m-k)^3$, where $m$ and $k$ are non-zero integers.

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- #1

- Feb 14, 2012

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Find all solutions to $(m^2+k)(m+k^2)=(m-k)^3$, where $m$ and $k$ are non-zero integers.

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- Feb 7, 2012

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Find all solutions to $(m^2+k)(m+k^2)=(m-k)^3$, where $m$ and $k$ are non-zero integers.

That gives three possible solutions, $(m,k) = (9,-21),\ (9,-6),\ (-1,-1)$.

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- Mar 5, 2012

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Why is that?therefore $m(m-8) = (m-4)^2 - 16$ must also be a square. The only way that can happen is if $m-4 = \pm5$.

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- Feb 7, 2012

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If $x^2-16 = y^2$ then $16 = x^2-y^2 = (x+y)(x-y)$. The only factorisation of $16$ giving positive integer values for $x$ and $y$ is $x+y=8$, $x-y=2$, so that $x=5$ and $y=3$.Why is that?therefore $m(m-8) = (m-4)^2 - 16$ must also be a square. The only way that can happen is if $m-4 = \pm5$.

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- #5

- Feb 14, 2012

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My solution:

By expanding and simplifying the given equation, we get

\(\displaystyle m^3+m^2k^2+mk+k^3=m^3-3m^2k+3mk^2-k^3\)

\(\displaystyle 2k^3+(m^2-3m)k^2+mk+3m^2k=0\)

Since $k \ne 0$, divide through the equation above by $k$, we have

\(\displaystyle 2k^2+(m^2-3m)k+m+3m^2=0\)

and solve for $k$ using the quadratic formula yields

\(\displaystyle k=\frac{-(m^2-3m) \pm \sqrt{(m^2-3m)^2-4(2)(m+3m^2)}}{2(2)}\)

\(\displaystyle \;\;\;=\frac{3m-m^2 \pm \sqrt{(m+1)^2(m(m-8))}}{4}\)

\(\displaystyle \;\;\;=\frac{3m-m^2 \pm (m+1)\sqrt{m(m-8)}}{4}\)

Recall that the expression inside the radical must be greater than or equal to zero, thus, \(\displaystyle m(m-8)\ge 0\) and this gives

$m\le 0$ and $m \ge 8$.

Also, bear in mind that we're told $m, k$ are both integer values, so, $m(m-8)$ has to be a squre.

But if we apply the AM-GM inequality to the terms $m$ and $m-8$, we find

\(\displaystyle \frac{m+m-8}{2} \ge \sqrt{m(m-8)}\)

\(\displaystyle m-4 \ge \sqrt{m(m-8)}\)

$m\ge 8$ implies $m-4 \ge 4$ and this further implies $\sqrt{m(m-8)} \le 4$ and now, we will solve for $m$ by considering $\sqrt{m(m-8)}=0$ or $1$ or $2$ or $3$or $4$ and we will take whichever solution that gives the integer values of $m$.

\(\displaystyle \sqrt{m(m-8)}= 4\) gives irrational $m$ values.

\(\displaystyle \sqrt{m(m-8)}= 3\) gives $m=-1, 9$.

\(\displaystyle \sqrt{m(m-8)}= 2\) gives irrational $m$ values.

\(\displaystyle \sqrt{m(m-8)}= 1\) gives irrational $m$ values.

\(\displaystyle \sqrt{m(m-8)}= 0\) gives irrational $m=0, 8$ values but $m>0$, hence $m=8$.

Thus, by substituting each of the value of $m$ to the equation (*) above to find for its corresponding $k$ value, we end up with the following 4 solution sets, namely

$(m, k)=(-1, -1), (8, -10), (9, -6), (9, -21)$

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- #6

- Feb 7, 2012

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Oops, I should have takenWhy is that?therefore $m(m-8) = (m-4)^2 - 16$ must also be a square. The only way that can happen is if $m-4 = \pm5$.