# What are the integer solutions of x4+y4=z4?

In many of our posts in the Proofs category, we have seen situation where attempting the impossible has inspired many people to come with truly beautiful results.  Another one of these problems that was presented to the world was, “are there any integer solutions to the equation $$x^{n}+y^{n}=z^{n}$$?”  This is Fermat’s Last Theorem, and while presented in the 17th century, it was only around 20 years ago that a proof of the theorem was published by Andrew Wiles.  While I will not attempt to show the intricacies involved the general proof, I did want to look at the case that Fermat himself proved, $$n=4$$.

### $$x^{4}+y^{4}=z^{4}$$ has no integer solutions.

We have most already shown nearly everything we need for the proof of Fermat’s last theorem for $$n=4$$.  As we go along, I will remind you where you would have learned the important parts of this proof.  For the first part of the proof, we will assume there is a solution and simplify this solution so that we have one where the greatest common divisor of all three terms is 1.

#### Assume there is a solution, then find a reduced solution.

• To begin with, we will work with a proof by contradiction as we outlined in That’s Irrational.
• We suppose that there are integer solutions.  We’ll call these integers $$x_{0},y_{0}$$ and $$z_{0}$$, respectively.
• Since $$a^{4}=(-a)^{4}$$ for all real numbers, we would then have that if $$x_{1}=|x_{0}|, y_{1}=|y_{0}|$$ and $$z_{1}=|z_{0}|$$, then $$x_{1},y_{1}$$ and $$z_{1}$$ is another solution to the equation where all numbers are positive.
• Now, recall that we found that every natural number has a unique factorization in Fundamental Theorem of Arithmetic.
• Therefore, we can factor $$x_{1},y_{1}$$ and $$z_{1}$$ into products of primes.
• If there exists a prime that divides all 3 of these, we can factor this prime out and arrive at a new solution.
• By doing this repetitively, we arrive at a solution $$x_{2},y_{2}$$ and $$z_{2}$$ where no prime divides all three terms.
• Suppose that exactly 2 of these are divisible by some prime $$p$$, say $$x_{2}$$ and $$y_{2}$$.  Then this implies that \begin{align*} (x_{2})^{4}+(y_{2})^{4}&=(z_{2})^{4} \\ 0 &=(z_{2})^{4} \text{mod} p.\end{align*}
• We then get $$(z_{2})^{4}$$ is divisible by $$p$$.  However, you found in Those are some prime irreducibles, that this implies $$z_{2}$$ is divisible by $$p$$.
• Since we know not all three are divisible by $$p$$, we must have that no prime divides any pair of the numbers.

#### Which of these three can be even then?

• Now, suppose that $$x_{2}$$ and $$y_{2}$$ are odd, then these can be written as $$2k_{x}+1$$ and $$2k_{y}+1$$, respectively.  Furthermore, we can write $$z_{2}=2k_{z}$$.
• We then have that \begin{align*} (2k_{x}+1)^{2}+(2k_{y}+1)^{2}&=(2k_{z})^{2} \\ 4k_{x}^{2} +4k_{x}+1+4k_{y}^{2}+4k_{y}+1& =4k_{z}^{2} \\ 2& \equiv 0 \text{mod} 4.\end{align*}.
• Since this is not true, we must have $$z_{2}$$ be odd and one of the $$x_{2}$$ or $$y_{2}$$ be odd.  We will assume $$x_{2}$$ is odd, otherwise, we can switch the labeling of $$x_{2}$$ and $$y_{2}$$.

#### Now that we have which ones are even, we rewrite the problem.

• Now that we know that \begin{align*} (z_{2})^{4}-(x_{2})^{4}&=(y_{2})^{4} \\ ((z_{2})^{2}-(x_{2})^{2})((z_{2})^{2}+(x_{2})^{2})&=(y_{2})^{4}.\end{align*}
• Note that if a number divides both $$(z_{2})^{2}-(x_{2})^{2}$$ and $$(z_{2})^{2}-(x_{2})^{2}$$, then it would both the sum and difference of these.  Therefore, it would divide $$2(x_{2})^{2}$$ and $$2(z_{2})^{2}$$.
• If this number is an odd prime, then it does not divide 2.  Therefore, it divides $$x_{2}$$ and $$z_{2}$$.  However, we have chosen these so that they have no common divisors.  Hence, an odd prime cannot divide both of these terms.
• If 4 divides both of these terms, we then have that $$(x_{2})^{2}$$ is even, which contradicts that $$(x_{2})$$ is odd.
• Therefore, the greatest common divisor of $$(z_{2})^{2}-(x_{2})^{2}$$ and $$(z_{2})^{2}+(x_{2})^{2}$$ is 2.
• Since $$y_{2}$$ is even, we have that $$y_{2}$$ can be written as $$2y_{3}$$ for some integer $$y_{3}$$.
• Therefore, $$((z_{2})^{2}-(x_{2})^{2})*((z_{2})^{2}+(x_{2})^{2})=16(y_{3})^{4}$$.
• Since 16 divides the product of these two, this implies that one of them is divisible by 2 and the other is divisible 8 and the remaining terms in the product have no common divisors.
• We now know that $$((x_{2})^{2}-(z_{2})^{2})((x_{2})^{2}+(z_{2})^{2})=y_{2}^{4}$$, we must have one is divisible by 2 and a perfect 4th power, and the other is divisible by 8 and a perfect fourth power.  Otherwise, we would have a different prime dividing both terms.

#### Two more cases to look at.

##### Case 1
• Suppose that $$(z_{2})^{2}-(x_{2})^{2}=2m^{4}$$ and $$(z_{2})^{2}+(x_{2})^{2}=8n^{4}$$ for some integers $$n$$ and $$m$$ with the greatest common divisor being 1.
• Note that $$m \neq 0$$, otherwise, $$x_{2}=z_{2}$$.
• Therefore, $$2m^{4}+8n^{4}=2(z_{2})^{2}$$, so $$m^{4}+4n^{4}=(z_{2})^{2}$$.
• Furthermore, $$8n^{4}-2m^{4}=2(x_{2})^{2}$$, so $$4n^{4}-m^{4}=(x_{2})^{2}$$.
• We then have that $$4n^{4}=(z_{2})^{2}-m^{4}=(z_{2}-m^{2})(z_{2}+m^{2})$$.
• Since $$m$$ and $$n$$ have no common divisors, we also have $$m$$ and $$z$$ have no common divisors.
• By a similar argument as we had above, 2 is the greatest common divisor of $$(z_{2}-m^{2})$$ and $$(z_{2}+m^{2})$$.
• Therefore, $$z_{2}-m^{2}=2k^{4}$$ and $$z_{2}+m^{2}=2j^{4}$$ for some integers $$k$$ and $$j$$.
• We now have that $$2m^{2}=(z_{2}+m^{2})-(z_{2}-m^{2})=2k^{4}-2j^{4}$$.
• This then gives us $$m^{2}=k^{4}-j^{4}$$.
• Note that we have shown that if $$(y_{2})^{2}=(x_{2})^{4}-(z_{2})^{4}$$ has integer solutions, then we can find new integers $$m,k$$ and $$j$$ such that $$m^{2}=k^{4}-j^{4}$$.  Here, $$j^{4} \leq 2j^{4}=z_{2}+m^{2} < z_{2}$$, since $$m^{2} > 0$$.
• Since $$j^{4} < z_{2}$$ and $$z_{2}$$ is a positive integer, we must have $$j<z_{2}$$.
##### Case 2
• Suppose that $$(z_{2})^{2}-(x_{2})^{2}=8m^{4}$$ and $$(z_{2})^{2}+(x_{2})^{2}=2n^{4}$$ for some integers $$n$$ and $$m$$ with the greatest common divisor being 1.
• Note that $$n \neq 0$$, otherwise, $$x_{2}=z_{2}=0$$.
• Therefore, $$8m^{4}+2n^{4}=2(z_{2})^{2}$$, so $$4m^{4}+n^{4}=(z_{2})^{2}$$.
• Furthermore, $$2n^{4}-8m^{4}=2(x_{2})^{2}$$, so $$n^{4}-4m^{4}=(x_{2})^{2}$$.
• We then have that $$4m^{4}=(z)^{4}-(n_{2})^{2}=(z_{2}-n^{2})(z_{2}+n^{2})$$.
• Since $$m$$ and $$n$$ have no common divisors, we also have $$n$$ and $$z$$ have no common divisors.
• By a similar argument as we had above, 2 is the only common divisor of $$(z_{2}-n^{2})$$ and $$(z_{2}+n^{2})$$.
• Therefore, $$z_{2}+n^{2}=2k^{4}$$ and $$z_{2}-n^{2}=2j^{4}$$ for some integers $$k$$ and $$j$$.
• We now have that $$2n^{2}=(z_{2}+n^{2})-(z_{2}-n^{2})=2k^{4}-2j^{4}$$.
• This then gives us $$n^{2}=k^{4}-j^{4}$$.
• Furthermore $$j^{4} \leq 2j^{4} =z_{2}-n^{2} < z_{2}$$ since $$n^{2} > 0$$.
• Since $$j^{4} < z_{2}$$ and $$z_{2}$$ is a positive integer, we must have $$j<z_{2}$$.

• Note that we started with the assumption that $$x_{2},z_{2}$$ and $$(y_{2})^{2}$$ were integer solutions of the equation $$y^{2}=x^{4}-z^{4}$$.
• We then arrived at new integer solutions, either $$m,k$$ and $$j$$ or $$n,k$$ and $$j$$ with $$0<j < z$$.
• If we repeat this process, we will arrive at new solutions, with a positive integer smaller than $$j$$, say $$j_{1}$$.
• Then, continuing this, we get integer an integer $$j_{2}$$ with $$0 < j_{2} < j_{1} < z$$.
• Since we can continue indefinitely, this will lead to there be infinitely many integers less than $$z$$ and greater than 0.
• This is a contradiction, so we cannot have integer solutions to the equation $$y^{2}=x^{4}-z^{4}$$.
• This implies we cannot have integer solution to $$x^{4}+y^{4}=z^{4}$$, because such a solution would lead to a solution of the previous equation.

### Conclusion

In this post, we have used a lot of the techniques and proofs that we have been providing throughout our posts in the Proofs series.  All of this knowledge that we have obtained is being coalesced into something wonderful, as we can provide proofs for new results that we wouldn’t have even known the answer to before.  As we continue to find more and more answers, we have also found even more questions.

I hope you are enjoying our journey through this series.  If you are, please let me know by using the like button, or by sharing the post on Social Media.

This site uses Akismet to reduce spam. Learn how your comment data is processed.