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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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3 Answers 3

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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
    Commented Mar 31, 2019 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? $\endgroup$ Commented Mar 22, 2019 at 3:10
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Starting with any perfect square $(x^2:x\ge 5)$, you can find a triple, if it exists, where $A^2+B^2=x^4$ with a function that begins by solving Euclid's $C$ function for $n$. $$C=x^2=m^2+n^2\Rightarrow n=\sqrt{x^2-m^2}\text{ where }2\le m \lt x$$

To demonstrate we start with the smallest example with an existing $C$ that is a perfect square $(7,24,25)$ and we let $n=\sqrt{25-m^2}$ where $2\le m \lt x.$ (Note: $x\gt m$ because we want a natural number $n$. Whenever this function yields an integer and $m>n$ we have the $m,n$ we neet for a triple. Testing from $2\le m \lt 5$, we find

$$\sqrt{25-2^2}=\sqrt{21}\qquad \sqrt{25-3^2}=\sqrt{16}\qquad \sqrt{25-4^2}=\sqrt{9}=3$$

Note: only the last test yielded an integer $4=m>n=3$ which is what we need. $$f(4,3)\Rightarrow A=16-9=7\quad B=2*4*3=24\quad C=\sqrt{16+9}=5\qquad7^2+24^2=5^4$$

Some values of $x$ beget multiple triples such as where $65^2-52^2,56^2\text{, and }60^2$ yielded integers.

$$f(52,39)=(1183,4056,65)\quad f(56,33)=(2047,3696,65)\quad f(60,25)=(2975,3000,65)$$

The percentage of valid squares and total triples compared to $X$ varies with $X$. Here are sample counts of integers, valid squares, and triples (i,v,t) up to $x$: $$(5,1,1)\quad(10,2,2)\quad(50,18,20)\quad(100,42,52)\quad(500,268,386)\quad(1000,567,881)\quad(5000,3129,5681)\quad(10000,6449,12471)\quad(25000,16611,34835)\quad(50000,33882,75196)$$ Read: Between $1$ and $1000$ are $567$ integers that form $881$ triples where $A^2+C^2=x^4$.

Here is pseudocode for generating $A^2+B^2=C^4$ for all integers up to an input limit.

INPUT LIMIT

$X=1$ TO LIMIT

$\quad M=2$ TO $X-1$

$\quad \quad N=SQRT(X^2-M^2)$

$\quad \quad $IF $N=INT(N)$ AND $M>N$

$\quad \quad \quad $PRINT $M^2-N^2$ "," $2*M*N$ "," $X$

$\quad \quad $ENDIF

$\quad $NEXT M

NEXT X

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