Prime Power!

Conclusion .

The only solution is $a=2,b=6,c=10$ , therefore the sum of all possible values of $c$ is $10$ .

Proof .

First we consider a special case: $a = 2$ . In this case,

$2^b = c^2-b^2 = (c+b)(c-b)$

Therefore $c+b=2^p, c-b=2^q$ for some non-negative integers $p,q$ such that $p+q=b$ and $q\leq p-1$ . So, consider the following chain of inequalities:

$2(2p-1)\geq 2(p+q)=2b=2^p-2^q\geq 2^p-2^{p-1}=2^{p-1}.$

Then we get $2^{p-2}\leq 2p-1$ . One can check for $p=2,3,4,5$ that this holds. For $p\geq 6$ , notice that $2^x > x$ for any $x$ , we have $2^{p-2}=16\cdot 2^{p-6} \geq 16((p-6)+1)> 2(p-6)+11=2p-1$ .

Now we check the four cases of $p$ . First, we rearrange $2(p+q)=2^p-2^q$ as $2^q+2q = 2^p-2p$ . (Since $2^q+2q$ is strictly increasing, this allows us to check just a few small $q$ 's.)

1. If $p=2$ then $2^q+2q=2$ , but it contradicts with $q\geq 1$
2. If $p=3$ then $2^q+2q=6$ . This also has no solution.
3. $p=4$ , then $2^q+2q=8$ . This has exactly one solution $q=2$ .
4. $p=5$ , then $2^q+2q=22$ . When $q=3$ , the LHS=14; when $q=4$ , the LHS=24. No solution.

Therefore in this special case of $a=2$ , the only possible case is $c+b=2^4=16, c-b=2^2=4$ , which gives us $b=6, c=10$ . It is easy to check that this is indeed a solution.

Now we assume that $a$ is an odd prime (so $a\geq 3$ ), and prove that in this case there is no solution at all. Suppose that $b=a^mb_1$ , where $m$ is a non-negative integer and $b_1$ is not divisible by $a$ . So $a^b + a^{am} b_1^a = c^a$ . Then, since $b\geq a^m \geq am$ , we know that $a^{am}$ divides $a^b$ , and thus divides $c^a$ . Therefore $a^m$ divides $c$ . Suppose that $c=a^mc_1$ , then we have $a^b = a^{am}c_1^a - a^{am}b_1^a$ $a^{b-am} = c_1^a - b_1^a$

If $b=am$ then $b_1^a+1=c_1^a \geq (b_1+1)^a > b_1^a + 1$ , a contradiction. Therefore $b>am$ . So we know that $a | (b_1^a-c_1^a)$ . It is evident that $c_1>b_1$ , so let $d=c_1-b_1>0$ . By Fermat's Little Theorem one can prove that $a|d$ . Indeed, $c_1^a\equiv c_1, b_1^a\equiv b_1 ~ (mod~a)$ , so $d=c_1-b_1\equiv c_1^a-b_1^a\equiv 0~(mod~a)$ .

Before I proceed further, I'd like to mention a fact that is probably known by the reader: for any prime number $n$ and integer $1\leq k\leq n-1$ , the binomial coefficients $C_n^k$ is divisible by $n$ but not by $n^2$ . Actually this is readily seen in the formula $C_n^k = \frac{n!}{k!(n-k)!},$ where there is no factor $n$ in the denominator, and there is exactly one $n$ in the numerator.

Now, suppose that $d=a^tx$ , where $t\geq 1$ is an integer, and that $x$ is a positive integer not divisible by $a$ . We have

$a^b=c_1^a-b_1^a=(d+b_1)^a-b_1^a$ $= d^a + C_a^1d^{a-1}b_1+\cdots +C_a^{a-2}d^2b_1^{a-2}+ C_a^{a-1}db_1^{a-1}$

The last term is equal to $a^{t+1}xb_1^{a-1}$ , so it is divisible by $a^{t+1}$ but not by a higher power of $a$ . However, all of the first $a$ terms $d^a, C_a^1d^{a-1}b_1,\cdots, C_a^{a-2}d^2b_1^{a-2}$ are divisible by $a^{t+2}$ .

Indeed, we have $d^a = a^{ta}x^a$ , where $ta\geq 3t = t+2t \geq t+2$ ; and for $k=1 \cdots a-2$ , $C_a^{k}d^{a-k}b_1^{k}$ is divisible by $ad^{a-k}=a^{1+t(a-k)}x^{a-k}$ , where $1+t(a-k)\geq 1+2t = t+1+t \geq t+2$ .

Therefore, we get $a^b = a^{t+2}y+a^{t+1}z$ where $y,z$ are integers such that $y>0$ , and that $z$ is not divisible by $a$ . So $a^b = a^{t+1}(ay+z)$ , which is a contradiction since $ay+z$ is both $>1$ and not divisible by $a$ , thus cannot be a power of $a$ . In conclusion, there is no solution when $a$ is an odd prime.

We will solve the equation $a^b+b^a=c^a$ with the given conditions.

Firstly, we consider the case $a \ge 3, a$ is a odd prime. Since $a^b+b^a=c^a$ , or equivalently, $a^b=c^a-b^a$ , we obtain $b \equiv c \pmod{a}$ .

• If $b \equiv c \not\equiv 0 \pmod{a}$ , by using Lifting the Exponent Lemma, we have : $v_a(c^a-b^a) = v_a(a)+v_a(c-b)$ , which is equivalent to $v_a(c-b)=b-1$ . Since $\displaystyle c^a-b^a=(c-b) \cdot \sum_{i=0}^{a-1} c^ib^{a-1-i}$ , we duduce that $\displaystyle \sum_{i=0}^{a-1} c^ib^{a-1-i} = a$ , which can not occur as $\displaystyle \sum_{i=0}^{a-1} c^ib^{a-1-i} > \sum_{i=0}^{a-1} 1 = a$ .

• Now if $b \vdots a$ and $c \vdots a$ , put $b=b_1 \cdot a^k; c=c_1 \cdot a^l, l,k \ge 1; a \nmid b_1, a \nmid c_1$ . If $k>l$ , by dividing both side of the given equation by $a^l$ , we obtain $a^{b-la} = c_1^a-b_1^a \cdot a^{a(k-l)}$ , then $c_1 \vdots a$ , contradict with the assumption $a \nmid c_1$ . Similarly, we deduce that $k=l$ . Then we obtain the following equation : $a^{b-ka}=c_1^a-b_1^a$ . By similar arguments as above, we obtain a contradiction. Then we now have a "small conclusion" that $a$ can not greater than $2$ .

Now we consider the case that $a=2$ . The given equation now can be rewritten as $2^b=(c-b)(c+b)$ . Put $c+b=2^m, c-b=2^n, m>n\ge 1, m+n=b$ . If $n=1$ , then $c=b+2$ , we have $2b+2=2^m=2^{b-1}$ , giving no solution. Hence, $n>1$ . Therefore $2b=2^m-2^n=2^n(2^{m-n}-1) \vdots 4$ , then $b \vdots 2$ . Since $2b=2^m-2^n$ , we deduce that $2b > 2^{\frac{b}{2}+1}-2^{\frac{b}{2}-1} = 3 \cdot 2^{\frac{b}{2}-1}$ . Now we can give a bound for $b$ : $2 \le b \le 6$ . Then $b \in \{2,4,6\}$ . By testing each case, we obtain the only solution : $(a,b,c)=(2,6,10)$ . Therefore, the needed sum is equal to $10$ .

Given that a is a prime and since a appears in the exponent twice and the base once so let's try moving the two terms to one side and leave the one term on the other side. We get a^b = c^a - b^a After factorizing, a^b = (c-b) (c^(a-1) + ... + b^(a-1)) therefore c-b divides a^b

so let c-b = a^q and (c^(a-1)+...+b^(a-1)) = a^r and q<=r consider q<=r, a^q<=a^r which implies that a^q|a^r now sub back in, c-b divides (c^(a-1)+...+b^(a-1))

Let us take mod c-b, So that expression becomes b^(a-1) + b^(a-1) + ... + b^(a-1) because c = b mod c-b if you count there are a terms in the sum, it becomes a b^(a-1) (mod c-b) but this term is 0 mod c-b because it is divisible by c-b so a b^(a-1) = 0 (mod c-b).

assume q >= 2 then b^(a-1) is divisible by a which implies b is divisible by a because a is prime and so now all the exponents are divisible by a and hence by quoting fermat's last theorem, we know that a has to be 2.

Hence, c^2 - b^2 = 2^b in fact c-b = 2^q and c+b = 2^r so b = (2^r - 2^q)/2 but b = q+r

We can force a contradiction out of this since the first expression is big and the second one is small. Therefore, there are no solutions except small ones, i.e. (2^r - 2^q)/2 is at least 2^(r-2) q+r is at most 2r so 2^r <= 8r so r<=5 which means b = q+r is at most 9.

We have to put in 9 values into the RHS of c^2=2^b+b^2 and check. Checking shows that (a,b,c) = (2,6,10) is a solution and since it is the only solution, sum of all possible c is 10.

We have known about the LTE technique that if x,y,n are positive integers and p is an odd prime, p is not a divisor of x and y but p is a divisor of $x-y$ then $v_{p}(x^n-y^n)=v_{p}(a-b)+v_{p}(n)$

At first we get that $c>b>1$ .

If $a=2$ then $2^b=c^2-b^2$ . We infer that there exists 2 positive integers m,n such that $m+n=b , c+b=2^m, c-b=2^n(m>n)$ . We can easily get $m=4,n=2$ (Just prove that $m<6$ ) so that $(a,b,c)=(2,6,10)$

If a is not equal to 2, we have 2 case:

Case 1 : $(b,c)=1$ . We infer that a is not a divisor of both b and c. So by using the LTE technique we will get $v_{a}(c^a-b^a)=v_{a}(c-b)+1$ . Note that $v_{a}(a^b)=b$ and $v_{a}(c^a-b^a)=v_{a}(a^b)$ so $v_{a}(c-b)=b-1\rightarrow c-b=a^{b-1}$ . But we have the inequality $c^a-b^a>a(c-b)=a^b$ which lead to a contradiction.

Case 2 : If b and c have common divisors we can infer that one of those divisors must be a. Take $c=a^h.t_{1} , b=a^k.t_{2}$ ( a is not divisor of $t_{1}$ and $t_{2}$ ).

• If $h=k$ then $t_{1}^a-t_{2}^a=a^{b-ah}$ . Use the LTE and the inequality similarly to Case 1 again and it makes another contradiction.

• if $h<k$ then we get $t_{1}^a-a^{a(k-h)}.t_{2}^a=a^{b-ah}$ .

We infer that $b=ah$ ( because a is not divisor of $t_{1}$ ). But $b=a^k.t_{2}>ah$ ( since k>h and b>1), again a contradiction.

• if $h>k$ , similarly we get $b=ak$ . But $b=a^k.t_{2}>ak$ ( since b>1), again a contradiction.

At the end we conclude that there is only $c=10$ that satisfies the equation.

a^(b-ma)=l^a-n^a=(l-n)(l^(a-1)+……+n^(a-1)), gcd of the two factors can only be 1 or a, so l-n=1 or l-n=a l-n=1 can not happen because l^a===l(mod a), n^a===n(mod a) if l-n=a and a is odd, then (l^(a-1)+……+n^(a-1))===a*n^(a-1) (mod a^2), which is not true so l-n=a=2 only solution is (a,b,c)=(2,6,10)

Generally, a rather sketchy solution. "So $a < b$ also leads to another contradiction" in particular is unjustified, looks like wishful thinking.

Let p = a. Note that c - b divides $c^a - b^a = p^b$ , so $c - b = p^k$ . Making these substitutions, we have $p^b + b^p = (b + p^k)^p$ . One notes that p must divide b, so we can let $b = mp$ . Making this substitution and dividing by $p^p$ on each side, one notes $m^p + (p^{m-1})^p = (m + p^{k-1})^p$ . By Fermat's Last Theorem, $p = 2 = a$ . Thus we are trying to solve $2^b + b^2 = c^2$ . This implies $2^b = (c + b)(c - b)$ . Each of $c + b$ and $c - b$ must be powers of $2$ . But by considering the difference between the two powers of $2$ , one notes $b < 11$ . We can now simply test every $b$ from $1$ through $10$ , and $b = 6$ is the only solution. This leads to $c = 10$ .

We first consider the case $a=2.$ Then $2^b=(c-b)(c+b)$ . So both $c-b$ and $c+b$ are powers of $2$ , with $c-b$ being strictly smaller: $c-b=2^u,$ $c+b=2^{u+v}$ , where $u$ and $v$ are positive integers. This implies $\begin{cases} 2^b=2^u2^{u+v}\\ 2b=2^{u+v}-2^u\end{cases}$ This system can be rewritten as $\begin{cases} 2u+v=b\\ b=2^{u-1}(2^v-1) \end{cases}$ So we get $2u+v=2^{u-1}(2^v-1).$ If $v\geq 2$ and $u\geq 4,$ then $2^v-1>v\geq 2$ and $2^{u-1}\geq 2u \geq 8.$ One can see that in this case the product of $2^v-1$ and $2^{u-1}$ is strictly greater than the sum of $v$ and $2u$ , so this is impossible. If $v=1,$ we get $2u+1=2^{u-1},$ which by parity implies $u=1$ , which does not work. If $u=1,$ then $2+v=2^v-1,$ which has no solutions. If $u=2,$ then $4+v=2(2^v-1),$ which has one solution: $v=2.$ This leads to $(a,b,c)=(2,6,10).$ Finally, if $u=3,$ then $6+v=4(2^v-1)$ , which has no solutions.

Suppose now that $a$ is an odd prime. By Fermat's theorem, $b^a\equiv b \pmod a$ and $c^a\equiv c \pmod a .$ Because $c^a-b^a=a^b\equiv 0 \pmod a ,$ we have $c\equiv b \pmod a.$

Suppose $c=b+a^ke,$ where $a$ and $e$ are coprime. Then $a^b=(b+a^ke)^a-b^a.$ Using the binomial formula, we get $a^b=ab^{a-1}a^ke+\frac{a(a-1)}{2}b^{a-2}a^{2k}e^2+...\equiv a^{k+1}b^{a-1}e \pmod{a^{k+2}}$ This implies that either $b=k+1,$ or $b\equiv 0 \pmod a.$ If $b=k+1,$ then $a^{k+1}=(c-b)(c^{a-1}+...+b^{a-1}),$ so $a^{k+1} > (c-b) \cdot a=a^{k+1}e,$ a contradiction.

So, $b$ must be divisible by $a.$ Suppose $b=ad$ . We can show that this is impossible in two different ways. The first way is to notice that $a^b+b^a=c^a$ can be rewritten as $(a^d)^a+b^a=c^a,$ which for odd $a$ has no solutions in positive integers by the Fermat's Last Theorem (proved by Andrew Wiles). The only problem with this argument is that it uses a very hard theorem, and there is a long-standing tradition in mathematics that strongly encourages the use of only those theorems for which you personally understand the proofs. So here is a different way to rule out this case, without using the "big guns".

The main idea is that the difference between two $a$ -th powers cannot be too small. Recall that if $b$ is divisible by $a$ , then so is $c$ . Suppose $c=af.$ Since $(a^d)^a+(ad)^a=(af)^a,$ we must have $(a^{d-1})^a+d^a=f^a.$ So $d^a\geq (a^{d-1}+1)^a-(a^{d-1})^a$ . By the binomial formula, this implies $d^a\geq a\cdot (a^{d-1})^{a-1}$ . This implies $d^a> (a^{d-1})^{a/2},$ so $d^2>a^{d-1}.$ If $a\geq 5$ , this implies $d^2>5^{d-1},$ which is only true if $d=1.$ If $a=3,$ we get $d=1,2,3,$ or $4.$

If $d=1,$ for any $a$ , then $b=a$ and we have $a^a+a^a=c^a.$ So $c^a=2a^a,$ which is impossible, because $2$ is not an $a$ -th power.

If $a=3$ and $d=2,$ then $a^b+b^a$ is $3^6+6^3=3^3\cdot 35,$ which is not a perfect cube. Similarly, $a=3$ and $d=3$ or $d=4$ do not work.

As a result, the only solution to our equation is $(a,b,c)=(2,6,10),$ so the answer is $10.$

If $a \ge 3$ , use the inequality, we can see that the equation can not be hold. If $a = 2$ , we have $2^b = (c-b)(c+b)$ and then we find out the unique solution of it is $b=6,c=10$ . So the sum of all positive integers $c$ satisfy the given equation is 10.