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It is well known that there exist integer solutions to the equation $a^2+b^2=c^2$. For example, an explicit formula for integer values of $a$ , $b$ , and $c$ is

\begin{align}a&=2mn \\ b&=m^2-n^2 \\ c&=m^2+n^2.\end{align}

I want to find solutions to the equation $a^2+b^2=c^4$. However, I got $a,b,c$ in a general form that substitutes $x$ in integer and get $a,b,c$ in the integer. What should I do first? I'm very stuck!

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Your equation, $a^2 + b^2 = c^4$, is the same as $a^2 + b^2 = c^2$, except that $c$ has been replaced with $c^2$. Thus you can find solutions by choosing $m$ and $n$ so that $$ \begin{align} a&=2mn \\ b&=m^2-n^2 \\ c^2&=m^2+n^2. \end{align} $$ But how can you make sure that $c$ is an integer? Easy: that last equation, $c^2 = m^2 + n^2$ says that $m,n$ and $c$ also form a Pythagorean triple. Thus you can find solutions with two new parameters, $r$ and $s$ such that $$ \begin{align} m&=2rs \\ n&=r^2-s^2 \\ c&=r^2+s^2. \end{align} $$ Substituting these in the first set of equations to find $a$ and $b$, you have $$ \begin{align} a&=4rs(r^2-s^2) \\ b&=6r^2 s^2-r^4-s^4 \\ c&=r^2 + s^2. \end{align} $$ For example, choosing $r=2$ and $s=1$ gives you $a = 24$, $b=7$, $c=5$ and, indeed, $24^2 + 7^2 = 5^4$.

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  • $\begingroup$ Finding any triplet where C is an odd square will satisfy the equation because the square root of C taken to the 4th power does is. $\endgroup$ – poetasis Mar 31 at 16:50
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Whenever you have $a^2+b^2=c^2$ you also get $(ca)^2+(cb)^2=c^4$ -- does that count for you?

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  • $\begingroup$ +1, but are those all solutions? $\endgroup$ – tarit goswami Mar 22 at 3:10

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