There exist no non-zero positive integer solutions to equation \(A^4+B^4=C^4\).
Proof:
We want to show the equation \(A^4+B^4=C^4\) has no non-zero integer solutions. We will proceed with proof by infinite descent, a technique created by Pierre de Fermat.
We begin by assuming, for the sake of contradiction, that such a solution to this equation does exist. We will denote it as \((a, b, c)\). We begin by rewriting the equation.
\[\begin{gathered} a^4 + b^4 = c^4 \\ \phantom{a^4} - b^4 \phantom{=} - b^4 \\ a^4 = c^4 - b^4 \\ (a^2)^2 = c^4 - b^4 \\ \end{gathered}\]
We will let \(z = a^2\), \(x = c\), and \(y = b\). Doing so, we can write this as the following equivalent equation:
\(x^4 - y^4 = z^2\)
For generalization, we will refer to this new equation as \(X^4-Y^4=Z^2\) where \((x,y,z)\) is a solution. To continue, we will factor the left hand side of the equation.
\[x^4 + y^4 = z^2\] \[(x^2+y^2)(x^2-y^2)=z^2\]
We can assume that \(x\) and \(y\) are coprime. But why? [Side Step 1]
Since \(x\) and \(y\) are coprime, this means the greatest common divisor, d, of \((x^2+y^2)\) and \((x^2-y^2)\) is either 1 or 2. But why? [Side Step 2]
Our proof will proceed by two cases. Case 1 will be \(d=2\), and Case 2 will be \(d=1\).
Case 1: \(d=2\)
If \(d=2\), then \(x\) and \(y\) are both odd while \(z\) is even. But why? [Side Step 3]
We know that \((y^2, z, x^2)\) form a primitive Pythagorean triple. But why? [Side Step 4]
As a result, we can make use of Euclid’s Formula. What is Euclid’s Formula and why does it work? [Side Step 5]
Through Euclid’s Formula, we can find two integers, \(d\) and \(e\), where \(gcd(d, e) = 1\) and \(d>e>0\) such that: \[z=2de\qquad y^2=d^2-e^2\qquad x^2=d^2+e^2\]
For the purpose of the proof, let’s multiply \(x^2\) by \(y^2\). We see: \[x^2\cdot y^2=\] \[(d^2+e^2)(d^2-e^2)=\] \[d^4-d^2e^2+d^2e^2-e^4=\] \[d^4-e^4\] Hence we have \(x^2y^2=d^4-e^4\). This creates a new solution to the equation \(Z^2=X^4-Y^4\), \((d,e,xy)\). But why? [Side Step 6]
The solution \((d,e,xy)\) to \(Z^2=X^4-Y^4\) has \(d<x\) from our original solution \((x,y,z)\). But why? [Side Step 7]
Hence, we have found a new solution, \((d,e,xy)\), starting from our original solution, \((x,y,z)\), where \(d<x\). If we were to repeat this process with our new solution, we could find yet another solution with a value less than \(d\).
Here we arrive at a contradiction. In using this process, we can find infinitely smaller and smaller solutions to the equation \(Z^2=X^4-Y^4\). However, this is not possible as there exists only a finite amount of small positive integer solutions. Therefore our original assumption that \((x,y,z)\) was a solution to \(Z^2=X^4-Y^4\) is wrong. Recall that \(x=c\), \(y=b\), and \(z=a^2\). It follows that no solutions \((a,b,c)\) exist to our original equation \(A^4+B^4=C^4\), as through substitution we see our original equation arise: \[z^2=x^4-y^4\] \[(a^2)^2=c^4-b^4\] \[+b^4\phantom{=}\phantom{(a^2)^2}+b^4\] \[a^4+b^4=c^4\] Therefore Case 1 is resolved.
Case 2: \(d=1\)
In this case, the \(gcd(x^2+y^2,x^2-y^2)=1\). For this to be true, \(x\) and \(y\) must be of different parity. But why? [Side Step 8]
We have \((x^2+y^2)(x^2-y^2)=z^2\). Since \((x^2+y^2)\) and \(x^2-y^2\) are coprime, and their product is a perfect square, each is also a perfect square respectively. But why? [Side Step 9]
Let \(s^2=x^2+y^2\) and \(t^2=x^2-y^2\). It stands that \(s\) and \(t\) must both be odd. But why? [Side Step 10]
Since \(s\) and \(t\) are both odd, their sum and difference are each even. But why? [Side Step 11]
Let us define two new variables, \(u\) and \(v\), such that: \[u = \frac{s+t}{2}\qquad v = \frac{s-t}{2}\]
It is safe to assume the sum and difference of \(s\) and \(t\) can be divided by two since we just showed that the sum and difference are both even, and any even number is divisible by 2.
It is also important to note that because the \(gcd(s^2,t^2)=1\), \(s\) and \(t\) are also both coprime. But why? [Side Step 12]
Since \(s\) and \(t\) are coprime, so are \(u\) and \(v\). But why? [Side Step 13]
It follows that at most, one of \(u\) and \(v\) can be even. But why? [Side Step 14]
We know that \(y^2=2uv\). But why? [Side Step 15]
Because \(y^2=2uv\), this implies exactly one of either \(u\) or \(v\) is even. But why? [Side Step 16]
Let’s assume \(u\) is even. We can write this as \(u=2m^2\) and \(v=k^2\). Now, we are able to find a solution set \((u,v,x)\) which forms a primitive Pythagorean triple. But why? [Side Step 17]
Through Euclid’s Formula, we can represent these values using smaller integers \(d\) and \(e\) where \(gcd(d,e)=1\) and \(0<d<x\). Doing so gives us: \[u=2de\qquad v=d^2-e^2\qquad x=d^2+e^2\] Since \(u=2m^2=2de\), and since \(gcd(d,e)=1\), both \(d\) and \(e\) must be perfect squares respectively. But why? [Side Step 18]
As a result, let us let \(d=g^2\) and \(e=h^2\). Note, \(g\leq d\) since \(g\) is a factor of \(d\) and factors of integers are always less than or equal to the integer itself. Also recall that \(v=d^2-e^2\)
Through substitution we see: \[v=\] \[d^2-e^2=\] \[(g^2)^2-(h^2)^2=\] \[g^4-h^4\]
Also recall that \(v=k^2\). Now it is possible to see: \[g^4-h^4=k^2\]
We have now found a smaller solution, \((g,h,k)\), where \(X=g\), \(Y=h\), and \(Z=k\). This solution is smaller than our starting one since \(0<g\leq d<x\), which means \(g<x\).
Using this new smaller solution, we can repeat this process to find yet another smaller solution, and do so infinitely. Here we arrive at a contradiction, since like the first case, this process should allow us to find infinitely smaller positive solutions, but there are only a finite amount of positive numbers. Hence, we are left to assume our original assumption is wrong, and that \((x,y,z)\) is not a solution to \(X^4-Y^4=Z^2\). It stands then that no solutions exist for our original equation either, as through substitution we see: \[z^2=x^4-y^4\] \[(a^2)^2=c^4-b^4\] \[+b^4\phantom{=}\phantom{(a^2)^2}+b^4\] \[a^4+b^4=c^4\]
Thus in having shown no solutions exist for either case, we can conclude that no non-zero positive integer solutions exist for the equation \(A^4+B^4=C^4\).
Q.E.D.