Is $100$ the only square number of the form $a^b+b^a$?

Question

Conjecture: $100$ is the only square number of the form $a^b+b^a$ for integers $b>a>1$.

In other words, $(a,b)=(2,6)$ is the only solution. Can we prove/disprove this?

Observations:

• The solution mentioned should not come as a surprise, since $2^6+6^2=8^2+6^2=10^2$ is a (non-primitive) Pythagorean triple. It is possible to show that $2^b+b^2$ has no other solutions. See Remark 1.

• In the general case where $a$ is a power of $2$; that is, $a=2^d$ for some positive integer $d$, a similar approach can be followed. See Remark 2.

• We can eliminate some values of $b$ when $a=5^r,6^r$, since no matter the value of $r$, we have $a\equiv5,6\pmod{10}$ respectively.

• PARI/GP code: If the conjecture is true it should only ever print 2 6.

  sqfun(a,b)={for(i=2,a,for(j=2,b,if(issquare(i^j+j^i)==1,print(i," ",j))));}
  

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Remark 1: Suppose that there is a positive integer $b$ that admits $2^b+b^2=t^2$ for some integer $t$. Then we can write the equation as $$2^b=(t+b)(t-b)\implies\begin{cases}t+b=2^c\\t-b=2^{b-c}\end{cases}$$ for some positive integer $c>\dfrac b2$. Subtracting the two equations yields $$2b=2^c-2^{b-c}\implies b=2^{b-c-1}(2^{2c-b}-1).$$ If $b$ is odd, it cannot have a factor of $2$, forcing $b-c-1=0\implies t=b+2$ and substituting gives $2^b+b^2=(b+2)^2$, or $2^b=4(b+1)$. No solutions exist.

If $b$ is even, then $b=2k$ for some positive integer $k$, so we must have $$\begin{cases}2^k=s(m^2-n^2)\\2k=2mns\end{cases}$$ for some integers $m,n,s$, so that $2^{mns}=s(m^2-n^2)$. Without loss of generality, let $m>n>0$ and $s>0$. [Servaes: If $mns\ge4$ then $2^{mns}\geq(mns)^2\geq sm^2> s(m^2-n^2)$, so the only solutions with even $b$ are $b=4,6$, and the first case does not yield a square.]

Remark 2: If $b$ is odd, it boils down to the equation $$2^{db}=4\left(b^{2^{d-1}}+1\right)\implies 2^{db-2}-1=b^{2^{d-1}}.$$ [Haran: For $db-2>1$, the LHS is congruent to $3\pmod4$, and since the RHS is a square for $d>1$, we reach a contradiction unless \begin{cases}d=1\implies a=2\quad\text{case covered above}\\db-2=1\implies1=b^{2^{d-1}}\implies 2^{d-1}=0\end{cases} which is impossible.]

If $b$ is even, then $b=2k$ for some positive integer $k$, and the Pythagorean triplet forces $$\begin{cases}2^{dk}=s(m^2-n^2)\\(2k)^{2^{d-1}}=2mns.\end{cases}$$ [Servaes: From the first equation, all three factors on the RHS are powers of two, so $$\begin{cases}m+n=2^u\\m-n=2^v\end{cases}\implies\begin{cases}m=2^{v-1}(2^{u-v}+1)\\n=2^{v-1}(2^{u-v}-1)\end{cases}$$ with $u>v>0$. Since $m$ and $n$ are coprime, we have $v=1$. Plugging this into the first equation yields $$2^{dk}=s(m-n)(m+n)=2^{u+1}s\implies s=2^{dk-u-1}.$$ Substituting this into the second equation yields $$(2k)^{2^{d-1}}=2mns=2(2^{u-1}+1)(2^{u-1}-1)s=(2^{2u-2}-1)2^{dk-u}$$ which is impossible; if we let $k=2^w\ell$ with $\ell$ odd then this implies $\ell^{2^{d-1}}=2^{2u-2}-1$ which by Catalan's conjecture/Mihailescu's theorem is impossible if $d>1$. Note that $u>v$ hence $u\geq2$.]

Answer 1

$\newcommand{\eps}{\varepsilon}$ $\newcommand{\rad}{\mathrm{rad}}$

At least, under the abc conjecture, there can be only finitely many pairs $(a,b)$ with $b>a>1$ coprime such that $a^b+b^a$ is a square.

As a reminder, the conjecture says that to any $\eps>0$ there corresponds some $K_\eps>0$ such that whenever $u,v$, and $w$ are coprime positive integers with $u+v=w$, one has $\rad(uvw)>K_\eps w^{1-\eps}$. Here $\rad(z)$ is the product of all primes dividing $z$ (thus, for instance, $\rad(8)=2$, $\rad(9)=3$, $\rad(10)=10$, $\rad(11)=11$, and $\rad(12)=6$).

Suppose now that $a^b+b^a=c^2$ with coprime integers $b>a\ge 3$ and $c>0$ (the case $a=2$ is resolved above). Applying the abc conjecture with $u=a^b$, $v=b^a$, $w=c^2$, and $\eps=0.05$, and making the key observation $\rad(a^bb^ac^2)\le abc$, we conclude that $$ Kc^{2\cdot 0.95} < abc $$ with an absolute constant $K>0$. At the same time, we have $c^2>a^b$ and $c^2>b^a$, implying $a<c^{2/b}$ and $b<c^{2/a}$, respectively. Consequently, $$ Kc^{1.9} < c^{(2/b)+(2/a)+1}, $$ showing that either $\frac1b+\frac1a>0.4$, or $Kc^{0.1}<1$. Clearly, there are only finitely many values of $c$ satisfying the latter condition, and to each value corresponds finitely many pairs $(a,b)$. On the other hand, since $\frac1b+\frac13\ge\frac1b+\frac1a>0.4$ implies $b<15$, there are only finitely many pairs $(a,b)$ satisfying the former condition. Thus, the total number of exceptional pairs $(a,b)$ is also finite.

Answer 2

(21-03-2020) Update: As no unconditional answer has yet been given, I will include a few more conditions that any solution must satisfy. The results used are rather advanced, so I will only include references, not proofs.

Lemma 5: Let $a$ and $b$ be positive integers such that $a^b+b^a$ is a perfect square. Then $\gcd(a,b)\leq3$.

Proof. Such a pair of positive integers yields a nontrivial integral solution to $$x^d+y^d=z^2,\tag{3}$$ where $d:=\gcd(a,b)$. This implies $d\leq3$ by this paper${}^1$.$\qquad\square$

Proposition 6: Let $a$ and $b$ be positive integers such that $a^b+b^a$ is a perfect square. Then $\gcd(a,b)\leq2$.

Proof. By the preceding lemma it suffices to show that $\gcd(a,b)=3$ is impossible. If $\gcd(a,b)=3$ then $a^{\tfrac b3}$ and $b^{\tfrac a3}$ are integers satisfying $$\Big(a^{\tfrac b3}\Big)^3+\Big(b^{\tfrac a3}\Big)^3=z^2,$$ for some integer $z$, which means that, after swapping $a$ and $b$ if necessary, either $$a^{\tfrac b3}=\frac{x(x^3-8y^3)}{w^2}z^2 \qquad\text{ and }\qquad b^{\tfrac a3}=\frac{4y(x^3+y^3)}{w^2}z^2,$$ for integers $x$, $y$ and $z$ with $x$ odd and $x$ and $y$ coprime, and $w:=\gcd(3,x+y)$, or $$a^{\tfrac b3}=\frac{x^4+6x^2y^2-3y^4}{w^2}z^2 \qquad\text{ and }\qquad b^{\tfrac a3}=\frac{3y^4+6x^2y^2-x^4}{w^2}z^2,$$ for $x$, $y$ and $z$ with $x$ and $y$ coprime and $x$ coprime to $3$, and $w=\gcd(2,x+1,y+1)$. These complete parametrizations of the integral solutions to $(3)$ when $d=3$ are taken from section 7.2 of this article${}^2$.

For the second parametrization, because $x$ is coprime to $3$ we see that $$\nu_3\Big(a^{\tfrac b3}\Big)=\nu_3(z^2) \qquad\text{ and }\qquad \nu_3\Big(b^{\tfrac a3}\Big)=\nu_3(z^2),$$ where $\nu_p(t)$ denotes largest integer $k$ such that $p^k$ divides $t$. It follows that $a\nu_3(b)=b\nu_3(a)$, and from $\gcd(a,b)=3$ it follows that either $\nu_3(a)=1$ or $\nu_3(b)=1$, so either $a$ divides $b$ or $b$ divides $a$, respectively. This means either $a=3$ or $b=3$.

If $a=3$ then the identity $$3^{\tfrac b3}=a^{\tfrac b3}=\frac{x^4+6x^2y^2-3y^4}{w^2}z^2,$$ shows that $z^2=3^{\tfrac b3}$. The parametrization for $b^{\tfrac a3}$ then shows that $3^{\tfrac b3}$ divides $b^{\tfrac a3}=b$, which quickly implies $b=3$. But $3^3+3^3=54$ is not a perfect square; a contradiction. If $b=3$ a similar argument shows that then $a=3$, and this shows that the second parametrization yields no solutions to our original problem.

For the first parametrization, note that $b$ is even and hence $a^{\tfrac b3}$ is a perfect square, hence so is $$w^2z^{-2}a^{\tfrac b3}=x(x^3-8y^3)=x^4-8xy^3.$$ This means there is some integer $c$ such that $x^4-8xy^3=(x^2-2c)^2$ and so $$cx^2+2y^3x-c^2=0.$$ In particular the discriminant $\Delta$ of this quadratic polynomial in $x$ is a perfect square, say $\Delta=(2e)^2$, where $$4e^2=\Delta=(2y^3)^2-4c(-c)^2=4(y^6+c^3).$$ It is a classical result that then for some nonnegative integer $k$ we have $$(|e|,c,|y|)=(3k^3,2k^2,k).$$ Plugging this in and solving the quadratic equation for $x$ yields $x\in\{\pm k,\pm2k\}$ where $y=\pm k$. Because $x$ and $y$ are coprime it follows that $k=1$, so $y=\pm1$ and $x\in\{\pm1,\pm2\}$. Then for $a^{\tfrac b3}$ to be a perfect square we must have $\{x,y\}=\{1,-1\}$, but then $b=0$, a contradiction. This shows that the second parametrization also yields no solutions to our original problem. In conclusion $\gcd(a,b)=3$ is impossible. $\quad\square$.

A small step towards a complete proof of the original problem would be to show that any solution other than $\{a,b\}=\{2,6\}$ must have $a$ and $b$ coprime, which seems likely. So for now, I propose the following conjecture:

Conjecture 7: Let $a$ and $b$ be positive integers such that $a^b+b^a$ is a perfect square and $\gcd(a,b)=2$. Then $\{a,b\}=\{2,6\}$.

Of course such solutions yield Pythagorean triples, for which the parametrization is well known. Perhaps arguments similar to those of Proposition 6 can be used here. I hope to give another update resolving this conjecture soon.

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References

1. H. Darmon and L. Merel, Winding quotients and some variants of Fermat’s last theorem, J.Reine Angew. Math. 490 (1997), 81–100.
2. H. Darmon, A. Granville, On the equations $x^p+y^q=z^r$ and $z^m=f(x, y)$, Bulletin of the London Math. Society, no 129, 27 part 6, November 1995, pp. 513–544.

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Original answer:

I'll collect a few partial results here. Let $a$, $b$ and $c$ be positive integers with $a,b>1$ such that $$a^b+b^a=c^2,$$ and let $d=\gcd(a,b)$. First two lemmas that are repeatedly useful.

Lemma 1: If $m$ and $n$ are positive integers with $m>n$ and not both even, such that $m+n$ and $m-n$ are both powers of $2$, then $m=2^k+1$ and $n=2^k-1$ for some positive integer $k$.

Proof. If $m+n=2^u$ and $m-n=2^v$ then $$m=\frac{(m+n)+(m-n)}2=\frac{2^u+2^v}2=2^{v-1}(2^{u-v}+1),$$ $$n=\frac{(m+n)-(m-n)}2=\frac{2^u-2^v}2=2^{v-1}(2^{u-v}-1),$$ and hence $v=1$ because one of $m$ and $n$ is odd. Then $k=u-v$.$\qquad\square$

Lemma 2: The only perfect power that is one less than a square is $8$.

Proof. There are fairly elementary proofs, but it also follows from Mihailescu’s theorem.$\qquad\square$

Proposition 3: If $a$ is a power of $2$ then $(a,b)=(2,6)$.

Most of this was proved in the original question by TheSimpliFire and Haran.

Proof. Let $a=2^e$. If $b$ is odd then writing $$(c-b^{2^{e-1}})(c+b^{2^{e-1}})=c^2-b^a=a^b=2^{be},$$ shows that both factors on the left hand side are powers of $2$. Then by Lemma 1 we have $c=2^v+1$ and $$b^{2^{e-1}}=2^v-1,$$ for some positive integer $v$ because $b$ is odd. Hence by Lemma 2 either $v=1$ or $2^{e-1}=1$. Clearly $v=1$ is impossible, so $2^{e-1}=1$ and so $e=1$. Then comparing exponents shows that $b=v+2$ and so $$v+2=b=2^v-1,$$ which is easily seen to have no integral solutions. Hence $b$ is even, say $b=2f$. Then we have the following Pythagorean triple: $$c^2=a^b+b^a=(2^e)^{2f}+(2f)^{2^e}=(2^{ef})^2+((2f)^{2^{e-1}})^2.$$ Then there exist positive integers $k$, $m$ and $n$ with $m>n$ and $\gcd(m,n)=1$ such that either $$c=k(m^2+n^2),\qquad2^{ef}=k(m^2-n^2),\qquad (2f)^{2^{e-1}}=2kmn,\tag{1}$$ $$\text{or}$$ $$c=k(m^2+n^2),\qquad2^{ef}=2kmn,\qquad (2f)^{2^{e-1}}=k(m^2-n^2).\tag{2}$$ In case the triple is of the form $(2)$, the middle identity shows that $k$, $m$ and $n$ are all powers of $2$, so in particular $n=1$ because $m$ and $n$ are coprime and $m>n$. Then the latter identity shows that $$(2f)^{2^{e-1}}=k(m^2-1)=k(m-1)(m+1),$$ where the factors $m-1$ and $m+1$ are odd and $k$ is a power of $2$, so both $m-1$ and $m+1$ are $2^{e-1}$-th powers. But for $e>1$ no two $2^{e-1}$-th powers of positive numbers differ by $2$, so $e=1$. Writing $k=2^u$ and $m=2^v$ we see that $u+v+1=f$, where $v\geq1$ because $m>n$. By comparing powers in the above we find that $$u+v+1=2^{u-1}(2^{2v}-1)=2^{u-1}(2^v-1)(2^v+1).$$ Of course $2^v+1>2$, so $2^{u-1}=1$ as otherwise $$2^{u-1}(2^v+1)>2^{u-1}+2^v+1\geq u+v+1,$$ a contradiction. Hence $u=1$ and $2^{2v}-1=v+2$, so also $v=1$. This yields the solution $(a,b)=(2,6)$.

On the other hand, if the Pythagorean triple is of the form $(1)$ then $k$, $m-n$ and $m+n$ are powers of $2$ because $$2^{ef}=k(m^2-n^2)=k(m-n)(m+n).$$ Because $m$ and $n$ are not both even, by Lemma 1 there exists a positive integer $v$ such that $m=2^v+1$ and $n=2^v-1$, and so the above shows that $k=2^{ef-v-2}$. Plugging this into the other equation yields $$(2f)^{2^{e-1}}=2kmn=2^{ef-v-1}(2^{2v}-1),$$ and writing $f=2^wg$ with $g$ odd then implies $$g^{2^{e-1}}=2^{2v}-1,$$ which by Lemma 2 implies that $2^{e-1}=1$, and hence $e=1$. Then $f=kmn$ and if $kmn\geq4$ then $$2^{kmn}\geq(kmn)^2\geq km^2>k(m^2-n^2),$$ so $f=kmn\leq3$. Then $b\leq6$ and clearly $b=2$ and $b=4$ do not yield solutions.$\qquad\square$

Proposition 4: If $d$ is even then $d=2$.

Proof. Suppose $d=2e$ and let $a=2eA$ and $b=2eB$. Then $e$ is odd as otherwise $c^2$ is the sum of two fourth powers, which is well known to be impossible by a classical result by Fermat. Now $$c^2=a^b+b^a=(a^{eB})^2+(b^{eA})^2,$$ is a pythagorean triple and hence there exist positive integers $k$, $m$ and $n$ with $m>n$ and $\gcd(m,n)=1$ such that $$a^{eB}=k(m^2-n^2),\qquad b^{eA}=2kmn,\qquad c=k(m^2+n^2),$$ where we can exchange the roles of $a$ and $b$ if necessary. It is clear that $$k=\gcd(a^{eB},b^{eA})=\gcd((dA)^B,(dB)^A)^e=d^{e\ell},$$ where $\ell\geq\min\{A,B\}$. In particular $k$ is an $e$-th power, and hence the factorizations $$(a^B)^e=k(m^2-n^2)=k(m-n)(m+n) \qquad\text{ and }\qquad (b^A)^e=2kmn,$$ show that, up to powers of $2$, the factors $m$, $n$, $m+n$ and $m-n$ are also $e$-th powers. That is to say, $$m=2^tp^e,\qquad n=2^uq^e,\qquad m+n=2^vr^e,\qquad m-n=2^ws^e,$$ for odd positive integers $p$, $q$, $r$ and $s$, and nonnegative integers $t$, $u$, $v$ and $w$, and $t+u+1\equiv v+w\equiv0\pmod{e}$. Then $$m=\frac{(m+n)+(m-n)}{2}=\frac{2^vr^e+2^ws^e}{2}=2^{v-1}r^e+2^{w-1}s^e,$$ $$n=\frac{(m+n)-(m-n)}{2}=\frac{2^vr^e-2^ws^e}{2}=2^{v-1}r^e-2^{w-1}s^e,$$ and at least one of $m$ and $n$ is odd, so either $v=1$ or $w=1$ (but not both) or $v=w=0$.

If either $v=1$ or $w=1$ (but not both) then $m$ and $n$ are both odd, so $t=u=0$ and hence $e=1$.

If $v=w=0$ then still either $t=0$ or $u=0$ because $m$ or $n$ is odd. If $u=0$ then $e\mid t+1$ and $$2^{t+1}p^e=2m=(m+n)+(m-n)=r^e+s^e,$$ and so it follows from Fermats last theorem that $e=1$. The same holds if $t=0$.$\qquad\square$