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

### 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))));}

• 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. – Servaes Jul 14 at 11:34
• @Servaes I have proven it for odd $b$ as well. – TheSimpliFire Jul 14 at 13:28
• 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... – Servaes Jul 16 at 1:40
• ...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 $u\geq2$. – Servaes Jul 16 at 1:40
• 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 $k$. – Servaes Jul 17 at 17:30

## 2 Answers

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

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 $$a and $$b, 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.

• You left out the relative primality condition in the description and use of the $abc$ conjecture. – KCd Jul 15 at 17:06
• @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). – W-t-P Jul 15 at 17:29
• Thanks. Regarding the coprimality condition, Beal's conjecture may help. – TheSimpliFire Jul 15 at 18:53

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 $$t=0$$.$$\qquad\square$$

• 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? – W-t-P Jul 17 at 18:56
• @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. – Servaes Jul 17 at 21:06
• 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. – TheSimpliFire Aug 2 at 9:24
• @TheSimpliFire It follows from the fact that $m$ and $n$ are coprime. – Servaes Aug 2 at 11:10