Great divisors or great powers

We first observe that since $(a,b)=1$ , we have $(a+b,a-b) = (a+b,2a) \mid 2(a+b,a) = 2(a,b) = 2$ . Thus, either $a+b$ and $a-b$ are relatively prime (in which case so are $(a+b)^{12}$ and $(a-b)^{61}$ ), or else they have gcd exactly 2 (and thus have only the prime 2 in common). The former case occurs precisely when one of $a,b$ is even, and the latter when $a,b$ are both odd.

Let's examine the "both odd" case in finer detail. Let $\nu(n$ ) denote the exponent of the largest power of 2 dividing $n$ . While $(a-b,a+b) = 2$ implies that one of $\nu(a-b)$ , $\nu(a+b)$ must be 1, the other one must be $\ge 2$ since $(a-b)(a+b) = a^2 - b^2 \equiv 1-1 \equiv 0 \pmod 8$ , so that $\nu(a-b) + \nu(a+b) \ge 3$ .

Let's assume for now that the pair $[\nu(a-b), \nu(a+b)]$ can actually attain any of the values deduced above, namely $[1,2], [1,3], [1,4], \ldots$ and $[2,1], [3,1], [4,1], \ldots$ . In that case, $((a+b)^{12},(a-b)^{61})$ will be equal to $2^c$ where $c = \min\{12,61r\}$ or $c = \min\{61,12r\}$ for some $r \ge 2$ . The former is always equal to $12$ , while the latter is one of $24,36,48,60,61$ . This gives six possible values for this case, plus a seventh for the "one odd, one even" case.

It only remains to verify the assumption about attainable pairs. For $[1,r]$ with $r\ge 2$ , take $a=1 , b = 2^r-1$ ; for $[r,1]$ with $r\ge 2$ take $a = 1, b=2^r+1$ .

(In the following solution, my motivation is in square brackets like [this].)

Consider a prime $p$ dividing both $a+b$ and $a-b.$ We show that either no such prime exists or $p = 2.$ [Testing small cases, I saw that the given GCD was often a power of $2,$ or $1.$ ] If there is no prime dividing $a+b$ and $a-b$ (for example, take $(a,b) = (6, 1),$ then the given GCD is clearly equal to $1.$

Otherwise, assume that such a $p$ exists. Then we have $\begin{cases} a+b&\equiv0\pmod p \quad (1) \\ a-b&\equiv 0\pmod p \quad (2) \end{cases}$ Adding these two congruences gives $2a \equiv 0 \pmod p.$ Thus, either $p = 2$ or $a \equiv 0 \pmod p.$ But if $a\equiv 0 \pmod p,$ then from equation $(2),$ we also have $b \equiv 0 \pmod p,$ contradicting the fact that $\gcd(a,b) = 1.$ Hence, $p=2.$

[I had confirmed my conjecture that the GCD was always a power of $2,$ but now I needed to find more information in order to bound the possible GCDs.] We now know that $a+b, a-b \equiv 0 \pmod 2.$ Since $a \equiv b \pmod 2$ and $\gcd(a,b) = 1,$ it must be that $a\equiv b \equiv 1 \pmod 2.$ Then, notice that the difference between $a+b$ and $a-b$ is $2b,$ and that $2b \equiv 2 \pmod 4.$ Since $a+b$ and $a-b$ differ by $2b$ which is not $0$ mod $4,$ they cannot both be $0$ mod $4$ - one must only contain one factor of $2.$ [This does it! Yes!] We have two cases:

1. $a+b$ contains only one factor of $2$

Then, $(a+b)^{12}$ contains exactly $12$ factors of $2,$ while $(a-b)^{12}$ contains more than $12$ factors, so the given GCD must equal $2^{12}.$

1. $a-b$ contains only one factor of $2$

Then, $(a-b)^{61}$ contains exactly $61$ factors of $2.$ Since the number of factors of $2$ in $(a+b)^{12}$ is a multiple of $12,$ the GCD must be in the set $\{2^{12}, 2^{24}, 2^{36}, 2^{48}, 2^{60}, 2^{61}\},$ where the last case occurs when $(a+b)^{12}$ has more than $61$ factors of $2.$

It remains to find examples that show that each of these values are achievable. After that, we may conclude that the GCD can take exactly $\boxed{7}$ values: $1, 2^{12}, 2^{24}, 2^{36}, 2^{48}, 2^{60},$ and $2^{61}.$

First of all, if a prime $p$ divides a power of a number then it divides the number itself. Therefore if $p | \gcd((a+b)^{12}, (a-b)^{61})$ then $p$ divides both $a+b$ and $a-b$ . But that means $p$ also divides $2a$ and $2b$ . Since $a$ and $b$ are coprime, the only possible $p$ is $2$ . Moreover, $4$ doesn't divide both $a+b$ and $a -b$ since $2$ would be a common factor of $a$ and $b$ .

This means that the only options with non-trivial $\gcd$ are ${\rm ord}(a+b) = k$ , ${\rm ord}(a-b) = 2$ with $k > 0$ (or the other way around but that doesn't contribute anything new to the $\gcd$ ). These option are realized by $a = 2^{k-1} + 1$ , $b = 2^{k-1} - 1$ for $k > 1$ and by $a=11, b=9$ for $k = 1$ . So the possible values of the $\gcd$ are $1, 2^{12}, 2^{24}, 2^{36}, 2^{48}, 2^{60}, 2^{61}$ and the answer is $\bf 7$ .

We can deal with powers of the two expressions later, and before that we will find the possible gcd of $(a+b), (a-b)$ .

Suppose that a positive integer $k \geq 1$ divides $(a-b)$ .

Keeping in mind the properties of modular arithmetic, we write:

$a-b \equiv 0 \pmod{k} \Rightarrow a \equiv b \pmod{k}$

Thus we can express $a$ in terms of $b$ and $k$ . $a=tk+b$ , where $t$ is some positive integer.

For $k$ to be a solution, we must have that $k \div (a+b)$ $(a+b)=(tk+2b)$

Thus, we must have $tk+2b \equiv 0 \pmod{k}$ $2b \equiv 0 \pmod{k}$ We note that $k$ cannot individually divide $a$ and $b$ (obviously exclusive of $k=1$ ), otherwise it'd not be a solution, since $gcd(a,b)=1$ . So, from the last equation we deduce that $k$ must be $2$ or $1$ . Thus, the solutions that exist for the original question can only be powers of $2$ .

$1$ is an obvious solution.

For all other solutions, both $a$ and $b$ must be odd. So, let $a=2t_1+1, b= 2t_2+1$ $gcd((a+b)^{12},(a-b)^{61})=gcd((2(t_1+1+t_2))^{12}),(2(t_1+t_2))^{61})$ . When $t_1+t_2$ is odd; $t_1+t_2+1$ is even, and vice-versa. Thus, we can have the following solutions : $1,2^{12i}$ where $1\leq i \leq 5, 2^{61}$ i.e a total of $\boxed{7}$ solutions.

To prove that no power $>61$ is a solution, we demonstrate the following:

First note that for a bigger solution to exist we must have that $a+b$ be a multiple of $2^m$ with $m>5$ .

$a \equiv -b \pmod{2^m}$ Let $a=p_1 \cdot 2^m -b$ where $p_1$ is a positive integer. $a-b=p_1 \cdot 2^m-2b=2(p_1 \cdot 2^{m-1} -b)$

Since $b$ has to be odd, $(p_1 \cdot 2^{m-1} -b)$ is also odd. So the maximum power of $2$ that we can extract from the above expression is $1$ . Of course we can have integers (a,b) such that $(a-b)^{61}$ contains a power of $2 > 61$ , but in that case, we use an analogous argument as used in this case, to conclude the same. Hence, no bigger solution exists.

Our answer: $\boxed{7}$

Let $d$ be the difference of $a$ and $b$ or $a-b$ since the numbers $a$ and $b$ do not matter as much as their difference and their sum. Note that the $gcd$ $\leq min( (a+b)^{12}, d^{61})$ .

Clearly, when $d=1$ , the $gcd = 1$ .

When $d \geq 2$ , since $a$ and $b$ are distinct coprime positive integers, both $a$ and $b$ are odd. Hence, their sum must be even. As such, when $d$ is odd, the $gcd$ will be $1$ . From this, we only have to consider $d$ which are even.

Note that when $d=2$ , the possible $gcd$ are $(2^{1})^{12}, (2^{2})^{12},(2^{3})^{12}, (2^{4})^{12},(2^{5})^{12}$ and $2^{61}$ .

When $d >2$ , $d$ can be a power of $2$ or a multiple of $2$ with 2 cases as follows:

Case 1: $d$ is a multiple of $2$ but not $4$ . Note that the maximum $gcd$ still remains at $2^{61}$ since $d$ only has one factor of $2$ . Hence, the possible $gcd$ for Case 1 is similar to the above when $d = 2$ .

Case 2: $d$ is a multiple of $2^{n}$ where $n \geq 2$ and is an integer. Note that the $gcd$ in this case cannot exceed $(2^{1})^{12}$ . This is because once the $gcd \geq (2^{2})^{12}$ , the sum of $a$ and $b$ must be a multiple of $4$ . Hence, the smaller number $b = \frac{a+b - d}{2} = \frac{2^{2}(\frac{a+b}{2^{2}} - \frac{d}{2^{2}})}{2} = 2(\frac{a+b}{2^{2}} - \frac{d}{2^{2}})$ , meaning that $b$ is even, which is false.

In conclusion the possible $gcd$ are $1, (2^{1})^{12}, (2^{2})^{12},(2^{3})^{12}, (2^{4})^{12},(2^{5})^{12}$ and $2^{61}$ . Hence, answer is $\boxed 7$ .