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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.

Case $a=2^d,\,d\in\Bbb N$ completed

Suppose that there is a positive integer $b$ that admits $$2^b+b^2=t^2\tag1$$ for some integer $t$. Then we can write the equation as $$2^b=(t+b)(t-b)\implies\begin{cases}t+b=2^c\\t-b=2^{b-c}\end{cases}\tag2$$ for some positive integer $c>\dfrac b2$. Subtracting the two equations yields $$2b=2^c-2^{b-c}\implies b=2^{b-c-1}(2^{2c-b}-1)\tag3.$$ If $b$ is odd, it cannot have a factor of $2$, forcing $$b-c-1=0\implies t-b=2\implies t=b+2\tag4$$ and substituting this into $(1)$ gives $$2^b+b^2=(b+2)^2\implies 2^b=4(b+1)\tag5$$ A calculus approach can be used by extending the domain of $b$ from $\Bbb N_{>1}$ to $\Bbb R$ and it is found that no solutions exist for odd $b$.

If $b$ is even, then $b=2k$ for some positive integer $k$, so we must have $$\begin{cases}2^k=s(m^2-n^2)\\2k=2mns\end{cases}\tag6$$ for some integers $m,n,s$, and we end up with the transcendental equation $$2^{mns}=s(m^2-n^2).\tag7$$

Credits to @Servaes: Without loss of generality, let $m>n>0$ and $s>0$. If $mns\ge4$ then $$2^{mns}\geq(mns)^2\geq sm^2> s(m^2-n^2),\tag8$$ so the only solutions with even $b$ are $b=4,6$, and the first case does not yield a square. $\square$


  • 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.

If $b$ is odd, it boils down to the equation $$2^{db}=4\left(b^{2^{d-1}}+1\right)\implies 2^{db-2}-1=b^{2^{d-1}}\tag9$$

Credits to @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\tag{10}\end{cases} which is impossible.

If $b$ is even, then $b=2k$ for some positive integer $k$, and the Pythagorean triplet forces $$\begin{cases}2^{dk}=s(m^2-n^2)\\(2k)^{2^{d-1}}=2mns\tag{11}.\end{cases}$$

Credits to @Servaes: From the first equation of $(11)$, all three factors on the RHS are powers of two, so $$\begin{cases}m+n=2^u\\m-n=2^v\tag{12}\end{cases}\implies\begin{cases}m=2^{v-1}(2^{u-v}+1)\\n=2^{v-1}(2^{u-v}-1)\end{cases}$$ with $u>v>0$. Since $m$ and $n$ are coprime, we have $v=1$. Plugging this into the first equation of $(11)$ yields $$2^{dk}=s(m-n)(m+n)=2^{u+1}s\implies s=2^{dk-u-1}.\tag{13}$$ Substituting $(13)$ into the second equation of $(11)$ yields $$(2k)^{2^{d-1}}=2mns=2(2^{u-1}+1)(2^{u-1}-1)s=(2^{2u-2}-1)2^{dk-u}\tag{14}$$ which is impossible; if we let $k=2^w\ell$ with $\ell$ odd then this implies $$\ell^{2^{d-1}}=2^{2u-2}-1\tag{15}$$ which by Catalan's conjecture/Mihailescu's theorem is impossible if $d>1$. Note that $u>v$ hence $u\geq2$.

  • 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. For the sake of illustration, consider $a=5$.

We know that an integer can never be a square number if its last digit is $2,3,7,8$. Therefore, $5^b+b^5$ is never square when $$5+b^5\equiv2,3,7,8\pmod{10}\implies b\equiv2,3,7,8\pmod{10}$$ This method can be extended to higher values of $r$.

  • As always, PARI/GP code below

(if the conjecture is true, it should only ever print out 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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    $\begingroup$ Since $b=2k=2mns>2$ the transcendental equation $$2^{mns}=s(m^2-n^2),$$ shows that without loss of generality $m>n>0$ and $s>0$. If $mns\geq4$ then $$2^{mns}\geq(mns)^2\geq sm^2> s(m^2-n^2),$$ so the only solutions with $b$ even must have $b\in\{4,6\}$, and $b=4$ does not yield a square. $\endgroup$ – Servaes Jul 14 at 11:34
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    $\begingroup$ @Servaes I have proven it for odd $b$ as well. $\endgroup$ – TheSimpliFire Jul 14 at 13:28
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    $\begingroup$ For the case $a=2^d$ and $b=2k$ you have found $$2^{dk}=s(m^2-n^2)=s(m-n)(m+n),$$ so all three factors on the RHS are powers of two, so $$m+n=2^u\qquad\text{and}\qquad m-n=2^v,$$ for nonnegative integers $u$ and $v$, and clearly $u>v>0$. Then $$m=\frac{(m+n)+(m-n)}2=\frac{2^u+2^v}2=2^{v-1}(2^{u-v}+1),$$ $$n=\frac{(m+n)-(m-n)}2=\frac{2^u-2^v}2=2^{v-1}(2^{u-v}-1),$$ but of course $m$ and $n$ are coprime, so $v=1$. (Continued... $\endgroup$ – Servaes Jul 16 at 1:40
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    $\begingroup$ ...Continued) Plugging this into the original equation yields $$2^{dk}=s(m-n)(m+n)=2^{u+1}s,$$ or equivalently $s=2^{dk-u-1}$. Then plugging this into your other 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 $k=2^w\ell$ 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 $u\geq2$. $\endgroup$ – Servaes Jul 16 at 1:40
  • $\begingroup$ This is the basic form of pythagorean triples; a primitive triple $a^2+b^2=c^2$ is of the form $$(a,b,c)=(m^2-n^2,2mn,m^2+n^2),$$ with $m$ and $n$ necessarily coprime. A general pythagorean triple is then a primitive one scaled by an integer factor $k$. $\endgroup$ – Servaes Jul 17 at 17:30
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$\newcommand{\eps}{\varepsilon}$ $\newcommand{\rad}{\mathrm{rad}}$

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.

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    $\begingroup$ You left out the relative primality condition in the description and use of the $abc$ conjecture. $\endgroup$ – KCd Jul 15 at 17:06
  • $\begingroup$ @KCd: thanks for bringing this to my attention, I have edited the answer (and will check whether this is a smarter way to deal with it). $\endgroup$ – W-t-P Jul 15 at 17:29
  • $\begingroup$ Thanks. Regarding the coprimality condition, Beal's conjecture may help. $\endgroup$ – TheSimpliFire Jul 15 at 18:53
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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 useful quite often.

Here are three lemmas I will use without further reference:

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: A perfect power is never one less than a square.

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

Proposition 1: 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 $a=2^d$ with $d>1$. If $b$ is odd then writing $$(c-b^{2^{d-1}})(c+b^{2^{d-1}})=c^2-b^a=a^b=2^{bd},$$ shows that both factors on the left hand side are powers of $2$. Then by Lemma 1 we have $c=2^v+1$ and $$b^{2^{d-1}}=2^v-1,$$ for some positive integer $v$ because $b$ is odd. Hence by Lemma 2 either $v=1$ or $2^{d-1}=1$. Clearly $v=1$ is impossible, so $2^{d-1}=1$ and so $d=1$. 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 $b=2e$. Then we have the following Pythagorean triple: $$c^2=a^b+b^a=(2^d)^{2e}+(2e)^{2^d}=(2^{de})^2+((2e)^{2^{d-1}})^2.$$ Then there exist positive integers $k$, $m$ and $n$ with $m>n$ and $\gcd(m,n)=1$ such that either $$c=k(m^2+n^2),\qquad2^{de}=k(m^2-n^2),\qquad (2e)^{2^{d-1}}=2kmn,\tag{1}$$ $$\text{or}$$ $$c=k(m^2+n^2),\qquad2^{de}=2kmn,\qquad (2e)^{2^{d-1}}=k(m^2-n^2).\tag{2}$$ 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 $$(2e)^{2^{d-1}}=k(m^2-1)=k(m-1)(m+1),$$ 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 $2^{d-1}$-th powers. But for $d>1$ no two $d$-th powers of positive numbers differ by $2$, so $d=1$. Writing $k=2^u$ and $m=2^v$ we see that $u+v+1=e$, 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 $(a,b)=6$.

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 $$2^{de}=k(m^2-n^2)=k(m-n)(m+n).$$ 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 $k=2^{de-v-2}$. Plugging this into the other equation yields $$(2e)^{2^{d-1}}=2kmn=2^{de-v-2}(2^{2v}-1),$$ and writing $e=2^wf$ with $f$ odd then implies $$f^{2^{d-1}}=2^{2v}-1,$$ which by Lemma 2 implies that $2^{d-1}=1$, and hence $d=1$. Then $e=kmn$ and if $kmn\geq4$ then $$2^{kmn}\geq(kmn)^2\geq km^2>k(m^2-n^2),$$ so $e=kmn\leq3$. Then $b\leq6$ and clearly $b=2$ and $b=4$ do not yield solutions.$\qquad\square$

Proposition 2: 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 $t=0$.$\qquad\square$

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  • $\begingroup$ In the proof of Proposition 2, you write: 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. Fermat theorem does not seem to apply as it refers to primitive solutions only, while $(2eA)^{eB/2}$ and $(2eB)^{eA/2}$ are not coprime, correct? $\endgroup$ – W-t-P Jul 17 at 18:56
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    $\begingroup$ @W-t-P I'm referring to Fermat's proof that the diophantine equation $$x^4+y^4=z^2,$$ has no solutions in the positive integers. There is no need to require $x$, $y$ and $z$ to be coprime; in fact if they are not, say $x=du$ and $y=dv$ with $\gcd(u,v)=1$, then $$z^2=x^4+y^4=(du)^4+(dv)^4=d^4(u^4+v^4),$$ and so $z=d^2w$ for some integer $w$ and $$u^4+v^4=w^2,$$ is a primitive solution. So if no primitive solutions exist, no (nonzero) solutions exist at all. $\endgroup$ – Servaes Jul 17 at 21:06
  • $\begingroup$ You write that $2mn=\left(\frac{b^A}{d^\ell}\right)^e$ implies that $m,n$ are $e$-th powers up to power of $2$. However, how about $2\cdot(2^1\cdot3^1)\cdot(2^1\cdot3^2)=6^3$, for example? Elaboration on this would be great. $\endgroup$ – TheSimpliFire Aug 2 at 9:24
  • $\begingroup$ @TheSimpliFire It follows from the fact that $m$ and $n$ are coprime. $\endgroup$ – Servaes Aug 2 at 11:10

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