# 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. 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))));}


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 $$u\geq2$$.]

• 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. Jul 14, 2019 at 11:34
• @Servaes I have proven it for odd $b$ as well. Jul 14, 2019 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... Jul 16, 2019 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 $u\geq2$. Jul 16, 2019 at 1:40


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, 2019 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). Jul 15, 2019 at 17:29

(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.

## 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.

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

• 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? Jul 17, 2019 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. Jul 17, 2019 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. Aug 2, 2019 at 9:24
• @TheSimpliFire It follows from the fact that $m$ and $n$ are coprime. Aug 2, 2019 at 11:10
• This is very nice; maybe even publishable. Mar 22, 2020 at 13:48