Question

$\frac{2h}{c} = t - \frac{1}{t^3}$ ①

$\frac{2k}{c} = \frac{1}{t} - t^3$ ②

Answer 1

c^2 (h^2 - k^2)^2 + 4h^3 k^3 = 0

Answer 2

Let the given equations be:

1. $\frac{2h}{c} = t - \frac{1}{t^3}$
2. $\frac{2k}{c} = \frac{1}{t} - t^3$

Let $A = \frac{2h}{c}$ and $B = \frac{2k}{c}$. The equations become:

$A = t - \frac{1}{t^3}$

$B = \frac{1}{t} - t^3$

We can rewrite the second equation as $B = \frac{t^2}{t^3} - t^3$. This doesn't look useful. Let's rewrite the second equation as $B = \frac{1}{t} - t^3$.

Consider multiplying the first equation by $t^3$:

$A t^3 = t^4 - 1$

Consider multiplying the second equation by $t$:

$B t = 1 - t^4$

From these two equations, we see that $A t^3 = -(B t)$.

$A t^3 + B t = 0$

$t(A t^2 + B) = 0$

Assuming $t \neq 0$ (otherwise the original expressions are undefined), we must have:

$A t^2 + B = 0$

Substitute back $A = \frac{2h}{c}$ and $B = \frac{2k}{c}$:

$\left(\frac{2h}{c}\right) t^2 + \left(\frac{2k}{c}\right) = 0$

$\frac{2}{c} (h t^2 + k) = 0$

Since $c \neq 0$, we have:

$h t^2 + k = 0$

If $h \neq 0$, we can write $t^2 = -\frac{k}{h}$.

Now substitute $t^2 = -k/h$ into the original equations. From equation ①:

$\frac{2h}{c} = t - \frac{1}{t^3} = t - \frac{1}{t \cdot t^2} = t - \frac{1}{t (-k/h)} = t + \frac{h}{kt}$

$\frac{2h}{c} = t + \frac{h}{kt}$

Multiply by $kt$:

$\frac{2hkt}{c} = kt^2 + h$

Substitute $t^2 = -k/h$:

$\frac{2hkt}{c} = k\left(-\frac{k}{h}\right) + h = -\frac{k^2}{h} + h = \frac{h^2 - k^2}{h}$

$\frac{2hkt}{c} = \frac{h^2 - k^2}{h}$

$2hk^2 t = c(h^2 - k^2)$

If $h \neq 0$ and $k \neq 0$, we can solve for $t$:

$t = \frac{c(h^2 - k^2)}{2hk^2}$

Square both sides:

$t^2 = \left(\frac{c(h^2 - k^2)}{2hk^2}\right)^2 = \frac{c^2 (h^2 - k^2)^2}{4h^2 k^4}$

We also have $t^2 = -\frac{k}{h}$. Equating the two expressions for $t^2$:

$-\frac{k}{h} = \frac{c^2 (h^2 - k^2)^2}{4h^2 k^4}$

Assuming $h \neq 0$ and $k \neq 0$, multiply both sides by $4h^2 k^4$:

$-k (4h k^4) = c^2 (h^2 - k^2)^2 h$

$-4h k^5 = c^2 h (h^2 - k^2)^2$

If $h \neq 0$, we can divide by $h$:

$-4 k^5 = c^2 (h^2 - k^2)^2$

This equation must hold if a value of $t$ exists.

Let's check the case $h=0$.

If $h=0$, the first equation becomes $\frac{2(0)}{c} = t - \frac{1}{t^3}$, so $0 = t - \frac{1}{t^3}$, which implies $t^4 = 1$. The second equation is $\frac{2k}{c} = \frac{1}{t} - t^3$. If $t^4 = 1$, then $t = \pm 1$ or $t = \pm i$.

If $t=1$, $\frac{2k}{c} = 1 - 1^3 = 0$, so $k=0$. If $t=-1$, $\frac{2k}{c} = \frac{1}{-1} - (-1)^3 = -1 - (-1) = 0$, so $k=0$. If $t=i$, $\frac{2k}{c} = \frac{1}{i} - i^3 = -i - (-i) = 0$, so $k=0$. If $t=-i$, $\frac{2k}{c} = \frac{1}{-i} - (-i)^3 = i - (i) = 0$, so $k=0$. So, if $h=0$, then $k=0$.

Let's check if $h=0, k=0$ satisfies the derived relation $-4 k^5 = c^2 (h^2 - k^2)^2$.

$-4 (0)^5 = c^2 (0^2 - 0^2)^2$

$0 = c^2 (0)^2$

$0 = 0$. So the relation holds for $h=0, k=0$.

The derived relation is $-4 k^5 = c^2 (h^2 - k^2)^2$. This can be rewritten as $c^2 (h^2 - k^2)^2 + 4 k^5 = 0$.

Let's verify this relation using the expression for $t$. We have $t^2 = -k/h$. Substitute this into the first equation:

$\frac{2h}{c} = t - \frac{1}{t^3} = t \left(1 - \frac{1}{t^4}\right)$

$\frac{2h}{c} = t \left(1 - \frac{1}{(t^2)^2}\right) = t \left(1 - \frac{1}{(-k/h)^2}\right) = t \left(1 - \frac{h^2}{k^2}\right) = t \left(\frac{k^2 - h^2}{k^2}\right)$

$\frac{2h}{c} = -t \left(\frac{h^2 - k^2}{k^2}\right)$

$t = -\frac{2h}{c} \frac{k^2}{h^2 - k^2} = \frac{2h k^2}{c (k^2 - h^2)}$

This matches the expression for $t$ we found earlier: $t = \frac{c(h^2 - k^2)}{2hk^2}$.

$\frac{2h k^2}{c (k^2 - h^2)} = \frac{c(h^2 - k^2)}{2hk^2}$

$(2h k^2)^2 = c^2 (k^2 - h^2)(h^2 - k^2)$

$4h^2 k^4 = c^2 (-(h^2 - k^2))(h^2 - k^2)$

$4h^2 k^4 = -c^2 (h^2 - k^2)^2$

$c^2 (h^2 - k^2)^2 + 4h^2 k^4 = 0$

Let's recheck the step $t^2 = -\frac{k}{h}$.

$h t^2 + k = 0$. This step is correct.

Let's look at the second equation:

$\frac{2k}{c} = \frac{1}{t} - t^3 = \frac{1}{t}(1 - t^4) = \frac{1}{t}(1 - (t^2)^2) = \frac{1}{t}(1 - (-k/h)^2) = \frac{1}{t}(1 - h^2/k^2) = \frac{1}{t}\left(\frac{k^2 - h^2}{k^2}\right)$

$\frac{2k}{c} = \frac{1}{t} \left(\frac{k^2 - h^2}{k^2}\right)$

$t = \frac{c}{2k} \left(\frac{k^2 - h^2}{k^2}\right) = \frac{c (k^2 - h^2)}{2k^3}$

Equating the two expressions for $t$:

$\frac{c(h^2 - k^2)}{2hk^2} = \frac{c (k^2 - h^2)}{2k^3}$

Assuming $c \neq 0$, $h \neq 0$, $k \neq 0$:

$\frac{h^2 - k^2}{h k^2} = \frac{k^2 - h^2}{k^3}$

$\frac{h^2 - k^2}{h k^2} = -\frac{h^2 - k^2}{k^3}$

$\frac{h^2 - k^2}{h k^2} + \frac{h^2 - k^2}{k^3} = 0$

$(h^2 - k^2) \left(\frac{1}{h k^2} + \frac{1}{k^3}\right) = 0$

$(h^2 - k^2) \left(\frac{k + h}{h k^3}\right) = 0$

This implies either $h^2 - k^2 = 0$ or $k+h=0$.

$h^2 = k^2$ or $h = -k$. If $h^2 = k^2$, then $h = \pm k$.

If $h = k$, then $h^2 - k^2 = 0$. If $h = -k$, then $h^2 - k^2 = (-k)^2 - k^2 = 0$.

So, if $h^2 = k^2$, the equation $(h^2 - k^2) \left(\frac{k + h}{h k^3}\right) = 0$ is satisfied.

In this case, $t^2 = -k/h = -k/(\pm k) = \mp 1$. If $h=k$, $t^2 = -1$. The equations become:

$\frac{2h}{c} = t - \frac{1}{t^3}$

$\frac{2h}{c} = \frac{1}{t} - t^3$

So $t - \frac{1}{t^3} = \frac{1}{t} - t^3$.

$t + t^3 = \frac{1}{t} + \frac{1}{t^3}$.

$t(1 + t^2) = \frac{1}{t^3}(t^2 + 1)$.

Since $t^2 = -1$, $t^2 + 1 = 0$. So $t(0) = \frac{1}{t^3}(0)$, which is $0=0$.

This means if $t^2 = -1$ (i.e., $t = \pm i$), then $t - 1/t^3 = 1/t - t^3$. If $t=i$, $i - 1/i^3 = i - (-1/i) = i + 1/i = i - i = 0$. $1/i - i^3 = -i - (-i) = 0$. So $\frac{2h}{c} = 0$, which means $h=0$. But we assumed $h=k$, so $k=0$.

This leads back to the $h=0, k=0$ case, which means $t^4=1$. But here we got $t^2=-1$, so $t^4=1$. This is consistent. So if $h=k$, the relationship is $h=k=0$.

If $h = -k$, then $t^2 = -k/(-k) = 1$. The equations become:

$\frac{2h}{c} = t - \frac{1}{t^3}$

$\frac{-2h}{c} = \frac{1}{t} - t^3$

So $\frac{2h}{c} + \frac{2h}{c} = (t - \frac{1}{t^3}) + (\frac{1}{t} - t^3) = t + \frac{1}{t} - (t^3 + \frac{1}{t^3})$.

$\frac{4h}{c} = t + \frac{1}{t} - (t^3 + \frac{1}{t^3})$.

If $t^2 = 1$, then $t = \pm 1$. If $t=1$, $\frac{4h}{c} = 1 + 1 - (1+1) = 0$, so $h=0$. Since $h=-k$, $k=0$. This is the $h=0, k=0$ case again. If $t=-1$, $\frac{4h}{c} = -1 + \frac{1}{-1} - ((-1)^3 + \frac{1}{(-1)^3}) = -1 - 1 - (-1 - 1) = -2 - (-2) = 0$, so $h=0$. Since $h=-k$, $k=0$. This is the $h=0, k=0$ case again.

The relation $c^2 (h^2 - k^2)^2 + 4h^2 k^4 = 0$ was derived assuming $h \neq 0, k \neq 0$. Let's look at the derived relation $c^2 (h^2 - k^2)^2 + 4 k^5 = 0$, derived from equating the expressions for $t^2$. This relation holds for $h \neq 0$. We checked that it holds for $h=0, k=0$.

What if $k=0$ and $h \neq 0$? If $k=0$, the second equation is $\frac{2(0)}{c} = \frac{1}{t} - t^3$, so $0 = \frac{1}{t} - t^3$, which means $t^4 = 1$. The first equation is $\frac{2h}{c} = t - \frac{1}{t^3}$. If $t^4 = 1$, then $\frac{1}{t^3} = \frac{t^4}{t^3} \frac{1}{t^3} = t$. So $\frac{2h}{c} = t - t = 0$, which means $h=0$. This contradicts the assumption $h \neq 0$. So, if $k=0$, we must have $h=0$.

The only case where the equations are satisfied is $h=0, k=0$ with $t^4=1$, unless there is a general relation between $h, k, c$ that holds for any valid $t$.

Let's re-examine $A t^2 + B = 0$.

$\frac{2h}{c} t^2 + \frac{2k}{c} = 0 \implies h t^2 + k = 0$.

If $h \neq 0$, $t^2 = -k/h$. We need another expression involving $t^2$ or $t^4$ or some power of $t$ that can be related to $-k/h$.

Consider $A t^3 = t^4 - 1$ and $B t = 1 - t^4$. $A t^3 = -(B t)$. $t(A t^2 + B) = 0$. $A t^2 + B = 0$ (assuming $t \neq 0$).

$\frac{2h}{c} t^2 + \frac{2k}{c} = 0$.

$h t^2 + k = 0$.

If $c^2(h^2 - k^2)^2 + 4h^3 k^3 = 0$, can we find a $t$?

If $h=0$, then $c^2(-k^2)^2 = 0 \implies c^2 k^4 = 0 \implies k=0$. If $h=k=0$, the original equations are $0 = t - 1/t^3$ and $0 = 1/t - t^3$, which means $t^4=1$. So any $t$ with $t^4=1$ works when $h=k=0$.

If $h \neq 0$, then $t^2 = -k/h$. The relation $c^2(h^2 - k^2)^2 = -4h^3 k^3$ implies that if $h$ and $k$ have the same sign (and are non-zero), then $h^3 k^3 > 0$, so $-4h^3 k^3 < 0$. Also $(h^2 - k^2)^2 \ge 0$ and $c^2 > 0$. So $c^2(h^2 - k^2)^2 \ge 0$. The equation $c^2(h^2 - k^2)^2 = -4h^3 k^3$ can only hold if both sides are zero. This happens if $h^2 - k^2 = 0$ AND $h^3 k^3 = 0$. $h^3 k^3 = 0$ implies $h=0$ or $k=0$. If $h=0$, then $k=0$ (from $h^2 - k^2 = 0$). If $k=0$, then $h=0$ (from $h^2 - k^2 = 0$). So, if $h$ and $k$ have the same sign (and are non-zero), the equation $c^2(h^2 - k^2)^2 + 4h^3 k^3 = 0$ only holds if $h=k=0$.

What if $h$ and $k$ have opposite signs? Let $k = -m$ where $m > 0$. Assume $h > 0$. $c^2(h^2 - (-m)^2)^2 + 4h^3 (-m)^3 = 0$ $c^2(h^2 - m^2)^2 - 4h^3 m^3 = 0$ $c^2(h^2 - m^2)^2 = 4h^3 m^3$. The left side is $\ge 0$. The right side is $> 0$ if $h > 0, m > 0$. In this case, $t^2 = -k/h = -(-m)/h = m/h > 0$. So $t$ is real.

The expression for $t$ derived from equation ① is $t = \frac{c(h^2 - k^2)}{2h^2 k}$. The expression for $t$ derived from equation ② is $t = \frac{c(h^2 - k^2)}{2kh^2}$. These two expressions for $t$ are identical.

The substitution of $t^2 = -k/h$ into the expression for $t$ should lead to an identity. $t = \frac{c(h^2 - k^2)}{2h^2 k}$. $t^2 = \frac{c^2(h^2 - k^2)^2}{4h^4 k^2}$. We must have $t^2 = -k/h$. $\frac{c^2(h^2 - k^2)^2}{4h^4 k^2} = -\frac{k}{h}$. $c^2(h^2 - k^2)^2 h = -k (4h^4 k^2)$. $c^2(h^2 - k^2)^2 h = -4h^4 k^3$. Assuming $h \neq 0$, $c^2(h^2 - k^2)^2 = -4h^3 k^3$.

$c^2(h^2 - k^2)^2 + 4h^3 k^3 = 0$.