Why does $p^{2e}|a^3$ give us $p^{3e}|a^3$? Anyways my solution: It is easy to eliminate $b=1$. Now write $k(a-1)=b-1$ for a positive integer $k$. Putting this back into the first condition gives $$(k(a-1)+1)^2|a^3 \Rightarrow (ka-k+1)^2|k^2 \cdot a^3$$Since $ka-ka+1$ and $k$ are coprime we can simplify by $k$ and get $(ka-k+1)^2|a^3$. For $k=1$, we have $a=b$ which is clearly satisfying the conditions. Assume that $k \neq 1$. Let the gcd of $k-1$ and $a$ be $d$ and $k=dx+1$, $a=dy$. $x$ and $y$ are coprime. Putting back into the last equation gives $((dx+1)dy-dx)^2|d^3y^3 \Rightarrow ((dx+1)y-x)^2|dy^3$. It is easy to see $(dx+1)y-x$ and $y$ are coprime. Thus simplifying $((dx+1)y-x)^2|d$. A simple bounding gives us that there is no other solution from here.

@above From $p^{2e}|a^3$ not follows that $p^{3e}|a^3$ For example $2^{2*3}|4^3$ but $2^{3*3} \not |4^3$

electrovector wrote: Why does $p^{2e}|a^3$ gives us $p^{3e}|a^3$? Uh I thought it was because all exponents of primes in $a^3$ are divisible by $3.$

We can easily see that (a,1) and (a,a) are solutions . We will show that there are no other solutions. Let d be the gcd of (a,b) so a=dx,b=dy,but then from the first condition we have that y^2 divides dx^3, but y^2 and x^3 are coprime so y^2 divides d. So d=y^2*s where s>=1 so a=y^2*s*x and b=y^3*s. From the second condition we have that y^2*s*x-1 divides y^3*s-1 . y^2*s*x-1 divides y^3*s*x-y and y^2*s*x-1 divides y^3*s*x -x so it divides their difference so y^2*s*x-1 divides x-y.If x=y then a=b,contradiction. So y^2*s*x-1<=|x-y| so |x-y|>=y^2*x-1. If y>x then y>y-x>=y^2*x-1 so 1>y*(xy-1) so xy=1 and x=y=1,false(y>x). If x>y we have that x-1>=x-y>=y^2*x-1 so 1>=y^2 so y=1. So from y^2*s*x-1<=x-1 we deduce that s<=1 so s=1 but then b=y^3*s=1,false. So the conclusion follows.

Let $a^3=kb^2$. Assume $b$ is not $1$ ($(a, 1)$ is solution). Then $a^3-1=kb^2-1$ so $a-1 | kb^2-1 \equiv k-1 (mod a-1)$. So either $k=1$ (then $a=x^2, b=x^3$ so $x^2-1 | x^3-1$ so taking modulo $x+1$, we see $x=1$), or $a \leq k$ so $a^3=kb^2 \geq ab^2$ so $a \geq b$ but $a-1 \leq b-1$ from the second divisibility so $a=b$.

Note that $(a,1)$ works for any $a$, assume $b>1$ and let $(a,b)=(da_1,db_1)$, where $a_1,b_1$ are coprime. Clearly $b_1\ge a_1$. We have $b_1^2\mid da_1^3$ and thus $d=\ell b_1^2$ for some $\ell$, yielding $\ell a_1b_1^2 -1 \mid \ell b_1^3 -1$. Consequently, $\ell a_1b_1^2-1\mid \ell b_1^2 (b_1-a_1)$ and therefore $\ell a_1b_1^2 -1 \mid b_1-a_1$. If $b_1\ne a_1$, then $\ell a_1b_1^2 +a_1\le b_1+1$ which holds only when $\ell=a_1=b_1=1$, a case we already ruled. Hence, $b_1=a_1$ and thus $(b,b)$ is the other solution family; concluding the problem.

Jalil_Huseynov wrote: Find all pairs $(a,b)$ of positive integers such that $a^3$ is multiple of $b^2$ and $b-1$ is multiple of $a-1$. Another bold action sol, during the test, i was just spamming stuff as i had like 1 hr remaining and i had no idea what to do, so i did prime factorization for bounding stuff, which supricingly ended up by working very well. Case 1: $b=1$ Then here $a$ can be literaly any integer so the pair $(a,b)=(a,1)$ works. Case 2: $b>1$ We divide in 2 subcases. Case 2.1: $a \ne b$ (i.e. we clearly have $b>a$) Let $b=\prod_{i=1}^{n} p_i^{a_i}$ and $N=\prod_{i=1}^{n} p_i^{\left\lceil \frac{2a_i}{3} \right\rceil}$, now using $b^2 \mid a^3$ we get that $v_{p_i}(a) \ge \left\lceil \frac{2a_i}{3} \right\rceil$ which means that for any $1 \le i \le n$ we get that $p_i^{\left\lceil \frac{2a_i}{3} \right\rceil} \mid a$ and using those along with the fact that all $p_i$'s are different we get that $N \mid a$, now from here we get that $N^3 \ge b^2$ and $a \ge N$, now since $\gcd(a-1,N)=\gcd(-1,N)=1$ (using euclidean alg) we use the second condition in this way, $a-1 \mid b-1$ means $a-1 \mid b-a$ which means that $a-1 \mid \frac{b}{N}-\frac{a}{N}$ so $a \left(\frac{N+1}{N} \right) \le \frac{b}{N}+1$ which means that $aN+a \le b+N \le b+a$ which gives $aN \le b$ but now this means that $N^4 \le (aN)^2 \le b^2 \le N^3$ so $N=1$ which means that $b=1$ which is a contradiction!. Case 2.2: $a=b$ All pairs with $a=b$ work. Hence all the pairs that work are $\boxed{(a,b)=(a,1), (a,a)}$ thus we are done

A quick solution. If $b=1$ then $a$ is anything. Otherwise, let $a=x+1$, then $b=kx+1$ for some $k>0$ and $x>0$. The condition is that $(kx+1)^2$ divides $(x+1)^3$. Consider the quotient of those numbers modulo $x$. Note that both $x+1$ and $kx+1$ are coprime to $x$, and both are $1$ modulo $x$, so their quotient is $1$ modulo $x$ as well. However, the quotient is obviously less than $x+1$ for all $k>1$. Thus, either $k=1$ which gives us $a=b$ which is another set of solutions, or the quotient equals $1$. We thus need to solve $(kx+1)^2=(x+1)^3$. Now obviously $x+1$ divides $kx+1$, which implies that $x+1$ divides $k-1$, which gives $k>x+1$. But then $(kx+1)^2\geq(x+1)^4$, impossible. Thus, there are no other solutions.

The answer is $a=b$ or $b=1$. Let $u=a-1$ and $tu=b-1$. Then we have $(tu+1)^2\mid (u+1)^3$. From now on assume $t\ge 2$. We have $(tu+1)^2\mid (t-1)^3$ as well by noting $(tu+1)^2 \mid \gcd(tu+1,u+1)^3$ Therefore, $(tu+1)^4\mid ((u+1)(t-1))^3$ We can see $tu+1\le 6$ or $u=0$ and bash cases.

Let $b-1 = z(a-1) \implies az - b = z-1$. First deal with the obvious cases, if $b = 1$, then $a$ can be any positive integer, if $z = 1$, then $a = b$ so assume all $a,b,z > 1$. Let $b = \prod_{i=1}^m p_i^{\alpha_i}$ and let $$N = \prod_{i=1}^m p_i^{\left \lceil \frac{2 \alpha_i}{3} \right \rceil}$$since $a^2 | b^3$, we have $N | a$ and so $N | az - b = z-1$. Since $z > 1$, we have $z \ge N+1$. So $b-1 = z(a-1) \ge (N+1)(N-1) = N^2 - 1$ so $b \ge N^2$. But since $2 \left \lceil \frac{2 \alpha_i}{3} \right \rceil > \alpha_i$ for $\alpha_i \ge 1$, we must have $b = 1$, but we assumed $b > 1$. So the only solutions are those claimed at the start - $(n,1)$ and $(n,n)$. $\blacksquare$

Obviously, if $a=1$ then $(a,b)=(1,1)$, and if $b=1$ then $(a,b)=(n,1)$. Suppose that $a,b>1$. Now, we may rewrite the problem statement to $$b^2\mid a^3\quad\text{and}\quad a-1\mid b-1.$$Let $a^3=kb^2$ where $k$ is positive integer. We have $$a-1\mid b-1\implies a-1\mid (b-1)(bk+k)=b^2k-k=a^3-k\implies a-1\mid k-1.$$Case 1: $k=1$ We have $a^3=b^2$. Thus, $a$ is a perfect square. Let $a=x^2$, so $b=x^3$ where $x>1$. Plug this into $a-1\mid b-1$. We have $$x^2-1\mid x^3-1\implies x+1\mid (x^2+x)+1\implies x+1\mid 1\implies x=0$$which contradict $x>1$. Case 2: $k>1$ Since $a-1\mid k-1$, there exists positive integer $m$ such that $$k-1=m(a-1)\implies k=am-m+1.$$Since $a-1\mid b-1$, we have $b\geq a$, and so $$a^3=b^2k=b^2(am-m+1)\geq a^2(am-m+1)\implies a\geq am-m+1\implies 0\geq (a-1)(m-1).$$Since $a>1$, we have $0\geq m-1$ so $m=1$ which give $a=b$. Clearly, $(a,b)=(n,n)$ work. $$\boxed{(a,b)=(n,1),(n,n)\text{ for every positive integers }n.}$$

If $b=1$ then any pair $(a,b)=(a,1)$ works henceforth assume $b>1$. Let $g=\gcd(a,b)$, write $a=gx$ and $b=gy$ with $\gcd(x,y)=1$. $g^2y^2 \mid g^3x^3 \implies y^2 \mid g$ since $\gcd(y^2,x^3)=1 \implies y \le \sqrt{g}$. $gx-1 \mid gy -1\implies gx-1 \mid g(y-x) \implies gx-1\mid y-x$ since $\gcd(gx-1,g)=1$. From here, $x(g+1) - 1 \le y$ or $x=y$. If $x\neq y$ then $$1(g+1) -1 \le x(g+1) - 1 \le y \le \sqrt{g} \implies g = 1$$but because $b>1$, two coprime numbers cannot divide each other which forces a contradiction. Thus $x=y$ which means $a=b$ and it is easy to verify that the pair $(a,a)$ works.

The solutions are $(n,1),(n,n)$ for all positive integers $n>1$(in all solutions above i haven't seen this criteria to be written) Assume $b \neq 1$. $a^3=mb^2$ for some $m \geq 1$. Clearly $$a-1 | a^3-1 \implies a-1 | mb^2-1$$Also we have $$a-1 | b-1 \implies a-1 | b^2-1$$So $$a-1 | (mb^2-1)-m(b^2-1) \implies a-1 | m-1 \implies a \le m$$So $$a^3=mb^2 \ge ab^2 \implies a \ge b$$$$a-1 | b-1 \implies a \le b$$Thus $a=b=n$ for all positive integers $n>1$ For b=1 any natural number $a > 1$ works.

Let $b = ka - (k-1)$. We have $ka - (k-1) | a^3$, so $$ka - (k-1) | -\left((ka)^3 - (k-1)^3\right) + k^3a^3 = (k-1)^3$$ However, we know $ka - (k-1)\ge a^{\frac{3}{2}}\implies k\le \sqrt{a} + 1$, so we have $$ka - (k-1) \ge (k-1)a\ge (k-1)^3$$ with equality only when $a = 1$ and $k = 2$, which is not a solution. Since $ka - (k-1) | (k-1)^3$, it follows that $k = 0$ or $k = 1$, which imply $b = 1$ and $b = a$, respectively. These both clearly work, so our solution set is $(a,1)$ and $(a,a)$ for all $a\in\mathbb{N}$ excluding $(1,1)$.

Jalil_Huseynov wrote: Find all pairs $(a,b)$ of positive integers such that $a^3$ is multiple of $b^2$ and $b-1$ is multiple of $a-1$. if $a,b>1$ let $d=(a,b)$ and $a=dx$,$b=dy$ then: $d^2y^2|d^3x^3\Rightarrow y^2|d\Rightarrow d=zy^2$ and $a-1|b-1\Rightarrow zxy^2-1|zy^3-1$ let $k(zxy^2-1)=zy^3-1$ then $zy^2|k-1\Rightarrow k=1,or,k\geqslant zy^2+1$ If$k\geqslant zy^2+1$ we have a contradiction so $k=1$. For $k=1$ we have $zxy^2-1=zy^3-1$ so $x=y=1$ we have the solotion $a=b$ and $b=1$

If $b=1$ then all pairs $(a,1)$ work, and if $a=b$ then all pairs $(a,b)$ work. Assume now that $a,b>1$ and $a \neq b$. We will show that no other solutions exist. Let $\gcd(a,b)=d$, and so $a=dk$ and $b=d\ell$ with $k,\ell$ being coprime. Then, the first condition rewrites as $\ell^2 \mid dk^3$, and so $\ell^2 \mid d$, thus we may write $d=\ell^2m$ with $m$ a positive integer,. The second condition rewrites as $(k\ell^2m-1) \mid \ell^3m-1$, and so $(k\ell^2m-1) \mid \ell^2m(\ell-k)$, and since $\gcd(k\ell^2m-1,\ell^2m)=1$, we infer that $(k\ell^2m-1) \mid \ell-k$. Since $a \neq b$ and $(a-1) \mid (b-1)$ we have $a <b$ and so $k < \ell$ too, hence $$\ell-1 \geq \ell -k \geq k\ell^2m-1 \geq k\ell^2-1 \geq \ell^2-1,$$and so $\ell=1$, absurd since $1=\ell>k \geq 1$. Thus, the only solutions are $(a,a)$ and $(a,1)$ with $a$ a positive integer.

nvm........

The answer are $(a,a)$ and $(a,1)$ If $b=1$, then it's trivial that all $a \in \mathbb Z_+$ works. Now we consider $b>1$. If $a\ne b$: Let $b=\prod{p_i^{k_i}}, a=\prod{p_i^{k'_i}}$. (the prime factorization) $a-1|b-1 \Rightarrow a-1|b-a$ $\Rightarrow a-1|\prod{p_i^{k'_i}}(\prod{p_i^{k_i-k'_i}}-1)$ $\Rightarrow a-1|\prod{p_i^{k_i-k'_i}}-1$ (cause $\gcd(a-1,\prod{p_i^{k'_i}})=1$) $\Rightarrow a-1|\prod{p_i^{k_i-k'_i}}-a$ $\Rightarrow a-1|\prod{p_i^{k_i-k'_i}}-\prod{p_i^{k'_i}}$ $\Rightarrow a-1|\prod{p_i^{k_i}}-\prod{p_i^{2k'_i}}=b-2a$ So, $a-1|b-2a-(b-a)=-a$, or $a-1|1 \Rightarrow a = 2$. Then we get $b=2$ or $b=1$, a contradiction! Hence $a=b$, which is another solution.

Clearly $a=b$ and $b=1$ works, now assume by contradiction we have a solution of nonequal positive integers that works, clearly $a<b$. Let the gcd of $a$ and $b$ be $d$. So let $dx=a$ and $dy=b$ for relatively prime $x$ and $y$. Then the first condition gives us $d^2y^2|d^3x^3$ so $y^2|dx^3$ meaning $y^2|d$ so let $d=y^2z$. Now the 2nd condition gives us $dx-1|dy-1$ so $dx-1|dy-dx$ so $dx-1|y-x$ now plugging in we get $xy^2z-1|y-x$ and this becomes the inequality $$xy^2z-1\leq y-x$$$$y^2z-1\leq y-1$$$$yz\leq 1$$Thus $y=1$ and $z=1$ thus $d=1$ thus $b=1$ but then $a<1$ contradiction. So only $a=b$ or $b=1$ works.

We have 2 cases. Case 1: $b = 1$. This means that $a$ could take on any value, hence our solution set for this case is $(a, 1)$. Case 2: $b \neq 1$: Since $b^2 | a^3$, say $a^3 = b^2k$. As $a - 1 | b - 1$, we have $b \equiv 1 \pmod {a - 1}$, as well as $a \equiv 1 \pmod{a - 1}$. Since $a^3 \equiv k \pmod{a - 1}$, $k \equiv 1 \pmod {a - 1}$. Therefore, $$a - 1 | k - 1$$Subcase 1: $k = 1$. This means that $a^3 = b^2$, implying $a$ and $b$ are perfect squares and perfect cubes respectively. Say $a, b = c^2, c^3$. $$\implies c^2 - 1 | c^3 - 1 \implies c^2 - 1 | c - 1$$This is impossible unless $c = 1$ which is excluded from this case as then $b = 1$. Subcase 2: $k \neq 1$. So, we have $k \geq a$. Since $b \geq a$ (as $a - 1 | b - 1$), we have: $$b^2k \geq ab^2 \geq a^3$$Since $a^3 = b^2k$, we have $b^2k = ab^2 \implies a = k \implies a^3 = b^3 \implies a = b$. So our solution set is $(a, a)$.

I don’t see my sol, so I’ll post it : ) We have $a-1\mid b-1$, that gives $b\equiv 1\pmod{a-1}$ and $b\geq a\vee b-1=0$; then we have \[a-1,\; b^2\mid a^3-b^2\]We now see that all couples $(a,1)$ work for $a>1$, then we assume $b\neq 1$: By $b \equiv 1\pmod{a-1}$ we get $1=gcd(b,a-1)=gcd(b^2, a-1)$ that gives \[ab^2-b^2\mid a^3-b^2\]that implies $a^3-b^2\geq ab^2-b^2\vee a^3-b^2=0$; the first case, combined with $a\leq b$ gives $a=b$ that works for all $a>1$, while the second gives \[a^{1\over 2}+1\mid 1\]that is impossible for $a>0$. Then, $(a,1), (a,a)$ for $a>1$ are the only solutions \qed

Our solutions are $\boxed{(k,k), (k,1) ~ \forall ~ k \ge 2}$. We can easily handle our edge cases $a=1,2$ and $b=1$. Otherwise, if $c = \tfrac{a^3}{b^2}$, we have \[a-1 \mid (a^3-1)-(b^2-1) = a^3 \left(\frac{c-1}{c}\right) \implies a-1 \mid c-1 \implies c \ge a.\] However, this implies $a \ge b$, and thus the second condition forces $a=b$. $\blacksquare$

Hi guys, can anyone check if the following solution is correct? **Step 1: Modulo Analysis and Factorization** Since $b-1$ is a multiple of $a-1$, we can write $b-1 = k(a-1)$ for some positive integer $k$. This implies $$b \equiv 1 \pmod{a-1}.$$ **Step 2: Inequality Method: Divisor Bounding** Since $a^3$ is a multiple of $b^2$, we have $b^2 \mid a^3$. By the divisor bounding inequality, we get $$b \le a^{\frac{3}{2}}.$$ **Step 3: Combining Modulo Analysis and Inequality** Combining the results from Step 1 and Step 2, we have: $$1 \le b \le a^{\frac{3}{2}} \text{ and } b \equiv 1 \pmod{a-1}.$$ This means $b$ must be one of the numbers $1, a, 2a-1, 3a-2, \dots$ up to the largest multiple of $a-1$ less than or equal to $a^{\frac{3}{2}}$. **Step 4: Casework** * **Case 1: b = 1** This case trivially satisfies the conditions for any positive integer $a$. * **Case 2: b = a** This case also trivially satisfies the conditions for any positive integer $a$. * **Case 3: b > a** In this case, we have $b = l(a-1) + 1$ for some positive integer $l \ge 2$. Since $b^2 \mid a^3$, we can write $$a^3 = m b^2 = m (l(a-1) + 1)^2$$for some positive integer $m$. Expanding the right side, we get a polynomial in $a$ with leading coefficient $ml^2$. Since the left side has leading coefficient $1$, we must have $ml^2 = 1$. This is impossible since $m$ and $l$ are both positive integers greater than 1. **Step 5: Product of Two Numbers is a Square and Four Numbers Theorem** We are left with the case where $a < b < 2a-1$. Since $b^2 \mid a^3$, we can write $a^3 = cb^2$ for some positive integer $c$. By the "Product of Two Numbers is a Square" technique, there exist positive integers $d, p, q$ with $\gcd(p,q) = 1$ such that $$a = dp^2 \text{ and } c = dq^2.$$ Substituting into $a^3 = cb^2$, we get $d^3p^6 = d^2q^2b^2$, or $dp^6 = q^2b^2$. By the Four Numbers Theorem, there exist positive integers $r, s, t, u$ such that $$d = rs, \ p^3 = tu, \ q = rt, \ b = su.$$ Since $\gcd(p,q) = 1$, we must have $\gcd(t,s) = 1$. Therefore, $s^2 \mid d$ and $t^2 \mid p^3$. This implies $s \mid r$ and $t \mid u$. Writing $r = vs$ and $u = wt$, we get $$d = v s^2, \ p^3 = vwt^3, \ q = vst, \ b = su.$$ Substituting into $b-1 = k(a-1)$, we get $$su - 1 = k(vs^2p^2 - 1) = k(vs^2vwt^3 - 1) = k(v^2s^2wt^3 - 1).$$ This implies $su - 1$ is divisible by $s^2$. However, since $s \ge 1$, we have $su - 1 \equiv -1 \pmod{s^2}$, a contradiction. **Conclusion** Therefore, the only solutions are $(a,b) = (a,1)$ and $(a,b) = (a,a)$ for any positive integer $a$.

Hi guys, can anyone check if the following solution is correct? Many thanks! **Step 1** Since $b-1$ is a multiple of $a-1$, we can write $b-1 = k(a-1)$ for some positive integer $k$. This implies $$b \equiv 1 \pmod{a-1}.$$ **Step 2** Since $a^3$ is a multiple of $b^2$, we have $b^2 \mid a^3$. By the divisor bounding inequality, we get $$b \le a^{\frac{3}{2}}.$$ **Step 3** Combining the results from Step 1 and Step 2, we have: $$1 \le b \le a^{\frac{3}{2}} \text{ and } b \equiv 1 \pmod{a-1}.$$ This means $b$ must be one of the numbers $1, a, 2a-1, 3a-2, \dots$ up to the largest multiple of $a-1$ less than or equal to $a^{\frac{3}{2}}$. **Step 4: Casework** * **Case 1: b = 1** This case trivially satisfies the conditions for any positive integer $a$. * **Case 2: b = a** This case also trivially satisfies the conditions for any positive integer $a$. * **Case 3: b > a** In this case, we have $b = l(a-1) + 1$ for some positive integer $l \ge 2$. Since $b^2 \mid a^3$, we can write $$a^3 = m b^2 = m (l(a-1) + 1)^2$$for some positive integer $m$. Expanding the right side, we get a polynomial in $a$ with leading coefficient $ml^2$. Since the left side has leading coefficient $1$, we must have $ml^2 = 1$. This is impossible since $m$ and $l$ are both positive integers greater than 1. **Step 5: Product of Two Numbers is a Square and Four Numbers Theorem** We are left with the case where $a < b < 2a-1$. Since $b^2 \mid a^3$, we can write $a^3 = cb^2$ for some positive integer $c$. By the "Product of Two Numbers is a Square" technique, there exist positive integers $d, p, q$ with $\gcd(p,q) = 1$ such that $$a = dp^2 \text{ and } c = dq^2.$$ Substituting into $a^3 = cb^2$, we get $d^3p^6 = d^2q^2b^2$, or $dp^6 = q^2b^2$. By the Four Numbers Theorem, there exist positive integers $r, s, t, u$ such that $$d = rs, \ p^3 = tu, \ q = rt, \ b = su.$$ Since $\gcd(p,q) = 1$, we must have $\gcd(t,s) = 1$. Therefore, $s^2 \mid d$ and $t^2 \mid p^3$. This implies $s \mid r$ and $t \mid u$. Writing $r = vs$ and $u = wt$, we get $$d = v s^2, \ p^3 = vwt^3, \ q = vst, \ b = su.$$ Substituting into $b-1 = k(a-1)$, we get $$su - 1 = k(vs^2p^2 - 1) = k(vs^2vwt^3 - 1) = k(v^2s^2wt^3 - 1).$$ This implies $su - 1$ is divisible by $s^2$. However, since $s \ge 1$, we have $su - 1 \equiv -1 \pmod{s^2}$, a contradiction. **Conclusion** Therefore, the only solutions are $(a,b) = (a,1)$ and $(a,b) = (a,a)$ for any positive integer $a$.

apmo 2022 p1 Let $k=\frac{a^3}{b^2}$ then we have that $a-1| k-1$ Which only happens when $a=b=k$ and hence we get that our solutions are $(n,n)$ and $(n,1)$.

Case 1: $b<a \implies b = 1, a > 1$, which always works. Case 2: $b = a$ also always works. Case 3: $b>a$. Let $a = dx, b = dy$ with $d = gcd(a, b)$. The equations reduce to $$y^2 \mid d, dx-1 \mid x-y.$$Now let $d = ky^2$. Then $$kxy^2 - 1 \mid x-y,$$so $$ky^2x-1 \le y-x \le xy-1$$$$\implies ky^2x \le xy \implies ky \le 1 \implies k = y = 1.$$But then $$b = dy = d \le dx = a,$$contradiction. So the only solutions are $\boxed{(a, b) = (t, 1), (t,t)}$ where $t \in \mathbb{N}$. $\square$

Here is a quick solution. We will show that the only solutions are $(a,b)=(a,1)$ and $(a,b)=(a,a)$. If $b=1$, $a$ can be any positive integer (apart from $1$) so suppose that $b\neq 1$. Let $a^3=k\cdot b^2$ where $k$ is a positive integer. Now look at this equation mod $a-1$ and keep in mind that $b\equiv 1 \pmod{a-1}$. We get $$k\equiv 1\pmod{a-1}$$ Now we begin our size argument. Case 1. $k=1$ We have that $a^3=b^2=k^6$ so $a=k^2$ and $b=k^3$. Now we have $$k^2-1\mid k^3-1\iff k+1\mid k^2+k+1\iff k+1\mid 1$$ So we have no solutions. Case 2. $k \ge a$ If $b=a$ we get a solution so suppose that $b\neq a$. We clearly have $a<b$ and $t=\frac{b-1}{a-1}\ge 2 \Rightarrow b \ge 2a-1$. Now we have $a^3=k\cdot b^2\ge a(2a-1)^2$. After expanding, this implies $$0\ge 3a^2-4a+1$$ And this is a contradiction. $\square$

If $b = 1$, we are done, so assume not. Let $a^3 = kb^2$, we have $a - 1 \mid a^3 - 1 = kb^2 - 1$, $a - 1 \mid b - 1 \mid kb^2 - kb$, so $a - 1 \mid kb - 1 \implies a - 1 \mid (k - 1)b$. Since $a - 1 \mid b - 1$, $(a - 1, b) = 1 \implies a - 1 \mid k - 1$. If $k = 1$, we have $a^3 = b^2$, so let $a = x^2$, $b = x^3$. $x^2 - 1 \mid x^3 - 1 \implies x + 1 \mid x^2 + x + 1$, a contradiction. If $k \geq a$, then $kb^2 > a^3$, unless $b = k = a$, which works, so we're done.

Clearly $b=1$ works, so assume $b>1$. Let $a^3=nb^2$ and $n=c^3d$ where $n,c,d\ge 1$ are integers and $c$ is maximized. Also note that $n\le b$ because $a\le b$. We see that we can let $a=cdk^2$ and $b=dk^3$ where $k\ge 1$ is an integer. Since $n\le b$, this means that $c\le k$. Plugging this in gives $(cdk^2-1)|(dk^3-1)$ so $(cdk^2-1)|(cdk^3-c)$ so $(cdk^2-1)|(k-c)$. Plugging in $k=c$ gives $a=b$, which works. When $c < k$, the LHS is bigger than the RHS. Therefore, the solutions are $\boxed{b=1}$ and $\boxed{a=b}$.

Solution by Mhremath & Sadigly $a-1\mid b-1$ Either $b=1$ or $b\geq a$. For $b=1$ case, $(a;b)=(a;1)$ obviously works Let $a^3=kb^2$ for a $k\in\mathbb{Z}^+$ Claim. $a-1\mid k-1$ Proof. We have $a-1\mid (b-1)(b+1)k$ or in other words $a-1\mid b^2k-k\Rightarrow a-1\mid a^3-k$. We also have $a-1\mid a^3-1$ for $a\neq1$. Subtracting these two gives us $a-1\mid k-1$ This could mean 3 things. 1. Case $k=1\Rightarrow a^3=b^2=x^6$ for some $x\in\mathbb{Z}^+/\{1\}$ $$a-1\mid b-1\Rightarrow x^2-1\mid x^3-1\Rightarrow x+1\mid x^2+x+1$$But we have $gcd(x+1;x^2+x+1)=1$ So this case can't be true 2. Case $k=a\Rightarrow a=b$ This obviously works. 3. Case $k>a\Rightarrow a^3=kb^2>ab^2\Rightarrow a>b$ But we have $b\geq a$, so this case can't be true,too

Let $k=a^3/b^2.$ \[0\equiv a^3-1 = kb^2-1\equiv k-1 \pmod{a-1}.\]However, $a-1\mid b-1\implies b\leq a \implies k\leq a.$ This forces $k=a,$ in which case $b^2=a^2\implies a=b$ or $k=1.$ When $k=1, a^3=b^2\implies (a,b)=(x^2,x^3).$ $$a-1\mid b-1 \iff x^2-1\mid x^3-1 \iff x+1\mid 1+x+x^2 \iff x+1\mid 1.$$Which is obviously impossible. Convsersely, We can easily verify that $(a,a)$ is a solution.

Taking $b=1$ works. Suppose $b>1$, then $a-1\mid b-1$ gives $a\leq b$. Note that $b^2(a-1) \mid a^3(b^2-1)$. So, $$b^2a-b^2 \mid -a^3(b^2-1)+(a^2+a+1)(b^2a-b^2)=a^3-b^2.$$Thus, $b^2a-b^2\leq a^3-b^2 \implies b\leq a$ (if $a^3=b^2$ then let $a=x^2$, $b=x^3$ and thus $x^2-1\mid x^3-1$ i.e. $x=1$, a contradiction) and it follows that $a=b$.