Wow. I did some crazy casework bash that was six pages long... *facepalm* EDIT: By the way, I got no solutions.

Or: So yeah a and b aren't divisible by 3. Cubic residues mod 9 are 0,1,8, hence $ab^5$ and $a^5b$ can either be 5 or 7 mod 9. But $a^5 \equiv a^{-1} \pmod{9}$ so we get linear systems of equations that in every case yield $3 | a$ contradiction.

Don't worry, I'm facepalming just as hard as you over my little mistake.

My solution was an entire page of a grid mod 9 of a, b, a^5*b+3, and a*b^5+3 for 21 cases where a and b aren't divisible by 3 and a mod 9 is less than or equal to b mod 9 WLOG.

We claim the answer is no. Assume the contrary. Let $x^3=a^5b+3$ and $y^3=ab^5+3$. Then, $x^3-y^3=ab(a-b)(a+b)(a^2+b^2)$. We claim that $3|ab(a-b)(a+b)(a^2+b^2)$. If $b$ is a multiple of 3, we are done. Else $a, a-b, a+b$ are distinct in mod 3 and so one is a multiple of 3. Thus, $3|x^3-y^3$. Note $n^3 \equiv -1, 0, 1 \pmod 9$, thus $x^3-y^3=-2, 0, 2 \pmod 9$, thus $x^3-y^3=0 \pmod 9$, thus $x \equiv y \pmod 3$. If $x \equiv y \equiv 0 \pmod 3$, then one of a,b is a multiple of 3. WLOG $3|a$. Then $3=x^3-a^5b$, but $27|x^3-a^5b$, fail. Thus, $a,b$ are not multiples of 3. Multiply these to get $a^6b^6+3ab(a^4+b^4)+9=(xy)^3$. But by Euler and FLT, we have $a^6b^6 \equiv 1 \pmod 9, a^4+b^4 \equiv 1^2+1^2 \equiv 2 \pmod 3$. Therefore, $3ab(a^4+b^4)+1 \equiv -1, 0, 1 \pmod 9 \implies 3ab(a^4+b^4) \equiv -2, -1, 0 \pmod 9 \implies 3ab(a^4+b^4) \equiv 0 \pmod 9$, but $a,b,a^4+b^4$ are not multiples of 3, contradiction. Therefore, no solutions.

No, otherwise take modulo $9$. Evidently $3 \nmid a,b$, so we get that $a^5b, ab^5 \in \left\{ 2,4 \right\} \pmod{9}$ since cubes are $\pm 1$ in this case. In other words, $ab^{-1}, ba^{-1} \in \left\{ 2,4 \right\} \pmod{9}$ by Euler, which is absurd since $2 \cdot 4 \equiv -1 \pmod 9$.

Just curious, did anyone try to find a and b that work?

No... The fifth powers kind of hinted at "no solution"

When I continuously failed, I tried for a little bit.

My logic: "they didn't ask to find all positive integers a and b so it must be false"

My solution, hopefully a 7 :\ Assume for the sake of contradiction that there are integers $a,b$ such that the given expressions are perfect cubes. Then, $a^5b, ab^5 \equiv 5,6,7 \pmod{9}$. However, by Euler's, $(ab)^6 \equiv 1 \pmod{9}$ when $ab$ is relatively prime to $9$. So, $(ab)^6 \equiv 0,1 \pmod{9}$. Considering the possible products for $a^5b, ab^5$, the only ones that pose a problem are when both are $6 \pmod{9}$. However, this is impossible (assume $3 \mid a$, contradiction, done), so there are no solutions. ??? EDIT: Sorry I actually put Euler, my bad. @djmathman Thanks, my bad for posting a repeat

It's actually Euler's, not FLT, but you should be fine.

The solution I posted earlier is exactly yours. I don't see why it shouldn't get the full 7 points.

oh shoot i put FLT

well i have a 6 now

Note that a cube must be $0,1,8 \pmod 9$. Thus, we must have $a^5b, ab^5 \equiv 5,6,7 \pmod 9$, but if either $a$ or $b$ is divisible by $3$, we have a contradiction. Thus, the only possible remainders are $5$ and $7$. Next, note that $a^6 \equiv 1 \pmod 9$ by Euler's Totient Theorem. Thus, $a^6b^6 \equiv 1 \pmod 9$. Taking cases on the remainders of $a^5b, ab^5$ and using this fact gives a contradiction for all cases, so there are no such integers $a$ and $b$.

Answer: No Proof: We start by considering $\mod 3$. Note that if either of $a,b$ is a multiple of 3, WLOG say $3 | a$, then $a^5b+3 \equiv 3 \mod 27$, which is a contradiction. Hence, $a,b$ are not divisible by $3$. Note that $n^3 \mod 9 \in \{1,-1\}$, which implies $a^5 b \equiv a^{-1} b \mod 9 \in \{5,7\}$ and similarly $b^{-1} a \mod 9 \in \{5,7\}$. However, notice that $a^{-1}b \cdot b^{-1} a \equiv 1 \mod 9$, implies that there exists two elements in $\{5,7\}$ which are inverses of each other. But simply checking yields a contradiction. QED.

IAmTheHazard wrote: I wrote a two-column proof for this as well. Link. I don't know why I did this. I'll leave you to decide what to make of it. bro just explained the reason for me to live my life for seeing further posts like this (Also, I am adding a few further lines on some ideas and stuff so that the mods don't consider my post as a spam) As done in some of the posts (and me myself, due to which I was stuck at the ending), you can observe that after getting $a\equiv 4\pmod{7}$ with $b\equiv\{5,6\}\pmod{7}$, you can just cycle the claim but this time for getting that $b\equiv 4\pmod{7}$ which instantly finishes the problem, which me being an idiot didn't even think about. And if you are wondering how I got these modulos, you can just product the two expressions and then some case bash $\pmod{7}$ gave me all this.

all i got is this Let $x^3=a^5b+3$ and $y^3=ab^5+3$. $x^3-y^3=ab(a-b)(a+b)(a^2+b^2)$, which is divisible by $3$

The answer is no. It's well known and easily proved that the cubic residues are -1,0,1 mod 9. Then, we want ab^5 and a^5b to be 5,6,7 mod 9. Now note that if a or b is a multiple of 3 the condition is not satisfied (and hence to be 6 mod 9 the number must be a multiple of 3, contradiction, so disregard that possibility). Multiply them together. We have by euler's totient that it's either 0 or 1 mod 9, however the product of 5x5, 5x7, or 7x7 does not yield a perfect cube.

really using mod 9 kills this problem

joshualiu315 wrote: really using mod 9 kills this problem yes because perfect cube -1,0,1 mod 9 or 7

Use $mod9$ to prove that this has no integers that satisfy this thing

Remark: The cracking insight is to make use of the fact that perfect cubes are either $-1, 0, 1 \bmod{9}$ and to utilize contradiction. We claim the answer is no, and shall prove the statement through contradiction. Suppose that $a^5b + 3 = p^3, ab^5 + 3 = q^3$ for integers $p, q$. Now note that any perfect cube $t$ satisfies one of the following conditions: $t \equiv -1 \bmod{9}$ or $t \equiv 0 \bmod{9}$ or $t \equiv 1 \bmod{9}$. Thus it must be true that either $a^5b \equiv ab^5 \equiv 5 \bmod{9}$ or $a^5b \equiv ab^5 \equiv 6 \bmod{9}$, or $a^5b \equiv ab^5 \equiv 7 \bmod{9}$. Now in all cases, $a^5b \equiv ab^5 \bmod{9} \implies a^5b - ab^5 \equiv 0 \bmod{9} \implies ab(a-b)(a^2 + b^2)(a+b) \equiv 0 \bmod{9}$, $a, b$ are relatively prime to $3$, and it cannot be true that $a \equiv b \bmod{9}$ or that $a + b \equiv 0 \bmod{9}$. These statements can be all be verified with properties of modular arithmetic. Thus $a^2 + b^2 \equiv 2 \bmod{3}$ which means $(a-b)(a+b) \equiv 0 \bmod{9}$ which will not be possible upon taking into consideration the conditions listed above, and analyzing quadratic residues modulo $9$.

Proof by contradiction If each of the 2 terms is a perfect cube, then their product must also be a perfect cube Easy to see that $a,b$ not divisible by 3 (take $\mod27$) Then take $\mod 9$ on their product to get that the product is $4,7(\mod9)$, so not a perfect cube So no solution exist Full proof here: https://infinityintegral.substack.com/p/usajmo-2013-contest-review

$a^5b+3=x^3$ $ab^5+3=y^3$ $ab(a^4-b^4)=x^3-y^3$ The LHS is always divisible by $3$, so $3|x^3-y^3$ $x^3 \equiv y^3 \pmod 3$ $x^3$ and $y^3$ can only be $-1,0,1$ mod $9$, so $x^3 \equiv y^3 \pmod 9$ $a^5b+3 \equiv ab^5+3 \pmod 9$ $a^5b \equiv ab^5 \pmod 9$ If $3|ab$, then one of $a,b$ divides $3$, so one of $ab^5+3$ or $a^5b+3 \equiv 3 \pmod 9$, which does not work, so $3\nmid ab$ $a^4 \equiv b^4 \pmod 9$ So $(a,b)$ can be $(1,8),(2,7),(4,5) \pmod 9$, or $a=b \pmod 9$ Plugging in all of them results in none working, so there are No Solutions

Assume for the sake of contradiction that this is possible. We break the problem up into cases as follows. Case 1: $3 \nmid a, b$. Consider mod 9. Notice that since cubes are $-1, 0, 1$ mod 9, we must have $a^5b+3 \equiv \pm 1 \pmod 3 \implies a^5b \equiv 5, 7 \pmod 3$. Similarly, we get that $ab^5 \equiv 5, 7 \pmod 3$. But by Euler's Theorem, we get that $a^6b^6 \equiv 1 \pmod 9$, which contradicts the above modular equalities. Case 2: $3 \mid a, b$. Again we use the fact that cubes can only be $-1, 0, 1$ mod 9. Now assume for the sake of contradiction that $3$ can divide $a$. We see that $a^5b+3 \equiv 3 \pmod 9$, contradiction. Similarly we see that $3 \nmid b$. Therefore, no such $a, b$ exist.

FTSOC, assume that $a$, $b$ s.t. problem conditions are satisfied exist. Note that neither $a$ or $b$ are multiples of $3$, otherwise (assuming WLOG $3\mid a$) \[\nu_3(ab^5+3)=\nu_3(243k+3)=1,\]which is definitely not a perfect cube. This also gives that the two numbers $a^5b+3$ and $ab^5+3$ aren't multiples of $3$. Now, note that since all cubes relatively prime to $9$ are $1$ or $-1$. Using this, we have that $a^5b$ and $ab^5$ must be $5$ or $7$ mod $9$ each, which gives that $a^6b^6$ can be $25$, $49$, or $35$ mod $9$. However, by Euler's totient function, we have that $a^6b^6$ for any $a$, $b$ relatively prime to $3$ must be $1$ mod $9$, a contradiction. Therefore no such $a$, $b$ exist, finishing the problem.

Please let me know if there are any errors; any pointers would be helpful. Also feel free to correct a math mistake, although I thought this proof works. We will let x=a^5+b and y=b^5+a in this proof First, we factor out the ab from the a^5*b and the b^5*a. Then, we divide by ab for both equations, giving us a^4=(x^3-3)/ab and b^4 (y^3-3)/ab. Raising both sides by 1/4 for both equations, we find that a=((x^3-3)/ab)^1/4 and b=((x^3-3)/ab)^1/4. Now, we multiply the first and second equations to find that ab=((y^3-3)(x^3-3)/(ab)^2)^1/4. Squaring and eliminating the ab term in the square root, we simply find that ab= ((x^3-1)(y^3-1))^1/2. Now, we will finish off the proof with algebra manipulation. We can sub in ab into both the first and second equations,(highlighted above for convenience+ease) to get a^2=((x^3-3)/((x^3-3)(y^3-3))^1/2)^1/4. Squaring inside the square root and eliminating, we find that a^2=((x^3-3)(y^3-3)) and that b^2=((y^3-3)(x^3-3)). That must mean that a=1/b. But, no pair of integers(yes, (1,1) technically works, but when subbing into the equation, you can easily find that it does not work) will satisfy that condition! Therefore, there are no integer values of (a,b) such that the problem statment is true. Remark I was surprised that there was no solution similar to mine- maybe I missed something? I understood why people would use mod for this problem, but I thought this problem was solvable via basic algebra manipulation. Anyways, again please let me know if you can see any obvious error in my proof- I am still working on proof-writing and am still not great.

The answer is no; FTSOC, assume otherwise, and let $(a,b)$ be a working pair. Clearly, $a$ and $b$ cannot be multiples of $3$. Consider the product of these two numbers: \[ a^6 b^6 + 3a^5 b + 3b^5 a + 9.\]Taking this modulo $9$, we have that $a^6 b^6$ is congruent to $1$. We have $a^5 b + b^5 a \equiv ab \pmod{3}$, so $3a^5 b + 3b^5a$ is congruent to either $3$ or $6$ mod $9$. So, the sum is congruent to either $4$ or $7$, neither of which are perfect cubes.