How do I solve this Diophantine equation, $a^{2n}+b^2=c^{2n}$, where $n$ is any postive integer $>1$ & $a,b,c\ne0$?
I tried it by applying the Pythagorean triplets generating formula, but unable to find any integer solutions & nor I'm able to prove that there are no integer solutions except $a,b,c=0$. Thanks in advance for any help or guidance.
Best Answer
Let $k>1$ be a positive integer, and let $a$, $b$ and $c$ be positive integers such that $$a^{2k}+b^2=c^{2k}.\tag{0}$$ Let $d:=\gcd(a,c)$ so that $d^{2n}$ divides $a^{2k}$ and $c^{2k}$, and hence also $b^2$. Then $d^k$ divides $b$ and so $(a,b,c)=(dA,d^kB,dC)$ for some positive integers $A$, $B$ and $C$ with $\gcd(A,C)=1$. It follows that also $\gcd(A,B)=\gcd(B,C)=1$, and it is easily verified that $$A^{2k}+B^2=C^{2k}.$$ So without loss of generality $a$, $b$ and $c$ are pairwise coprime. Then $(a^k,b,c^k)$ forms a primitive Pythagorean triple, and so either $$a^k=m^2-n^2,\qquad b=2mn,\qquad c^k=m^2+n^2,$$ or $$a^k=2mn,\qquad b=m^2-n^2,\qquad c^k=m^2+n^2,$$ for some coprime positive integers $m$ and $n$ with $mn$ even. In the first case we see that $$a^k+c^k=(m^2-n^2)+(m^2+n^2)=2m^2.$$ But by the main theorem of this article such a solution does not exist for $k\geq3$. In the second case we see that $$a^k+c^k=(2mn)+(m^2+n^2)=(m+n)^2.$$ Again by the main theorem of that article such a solution does not exist for $k\geq4$. So either $k=2$ or $k=3$.
For $k=2$ it is a classical result of Fermat that the Diophantine equation $$x^4-y^4=z^2,$$ has no nontrivial integral solutions. See also this question on math.stackexchange: Solving $x^4-y^4=z^2$ . This leaves only the case $k=3$, for which I do not see a slick solution.
Proposition: The Diophantine equation $x^6-y^6=z^2$ has no primitive integral solutions with $y\neq0$ even.
Proof. Let $x$, $y$ and $z$ be pairwise coprime integers such that $x^6-y^6=z^2$, with $y$ even and $|y|>0$ minimal. Then $(u,v,w)=(x^2,-y^2,z)$ satisfies $$u^3+v^3=w^2,$$ and it is a classical result by Euler that any primitive solution is of one of the following three forms: \begin{align} (1)&\begin{cases} u&=s(s+2t)(s^2-2st+4t^2)\\ v&=-4t(s-t)(s^2+st+t^2)\\ w&=\pm(s^2-2st-2t^2)(s^4+2s^3t+6s^2t^2-4st^3 + 4t^4) \end{cases} & 2\nmid s,\ 3\nmid(s-t),\\ \\ (2)&\begin{cases} u&=s^4-4s^3t-6s^2t^2-4st^3 + t^4\\ v&=2(s^4+2s^3t+2st^3+t^4)\\ w&=3(s-t)(s+t)(s^4+2s^3t+6s^2t^2+2st^3+t^4) \end{cases} &2\nmid(s-t),\ 3\nmid(s-t),\\ \\ (3)&\begin{cases} u&=-3s^4 + 6s^2t^2 + t^4\\ v&=3s^4+6s^2t^2-t^4\\ w&=6st(3s^4+t^4) \end{cases} &2\nmid(s-t),\ 3\nmid t, \end{align} after swapping $u$ and $v$ if necessary. In all three cases $s$ and $t$ are coprime integers. For a proof, see $\S 14.3.1$ of Henri's Cohen's Number Theory, Vol II, Analytic and Modern Tools.
Because $w=z$ is odd the solution $(u,v,w)=(x^2,-y^2,z)$ cannot be of the form $(3)$. It cannot be of the form $(2)$ because $-y^2\equiv0\pmod{4}$, whereas $u\equiv1\pmod{4}$ and $v\equiv2\pmod{4}$. It folows that it is of the form $(1)$, and so \begin{eqnarray} x^2&=&s(s+2t)(s^2-2st+4t^4)&=&s(s^3+8t^3),\\ y^2&=&4t(s-t)(s^2+st+t^2)&=&4t(s^3-t^3) \end{eqnarray} for two coprime integers $s$ and $t$ with $s$ odd and $3\nmid(s-t)$. The factors in each product are pairwise coprime, and hence they are all perfect squares, say $$s=S^2,\qquad t=T^2,\qquad s+2t=q^2,\qquad s^3-t^3=p^2,$$ for some integers $S$, $T$ , $p$ and $q$, not necessarily prime. Reducing mod $4$ shows that $t$ is even, and so we find $$S^6-T^6=p^2,$$ with $S$, $T$ and $p$ pairwise coprime and $T$ even, where clearly $|S|<|x|$ and $|T|<|y|$. This contradicts the minimality of $|y|$, and so we conclude that there does not exist any primitive solution with $y\neq0$ even.$\qquad\square$
Together this finally shows that equation $(0)$ does not have any solutions in positive integers $a$, $b$ and $c$ for any integer $k>1$.