Pair of multiples

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#1 h May 17, 2022, 6:44 PM • 12 Y

Y by megarnie, Iora, Mahmood.sy, ImSh95, PHSH, S.Ragnork1729, farhad.fritl, Mango247, Mango247, Mango247, rudransh61, ItsBesi

Find all pairs $(a,b)$ of positive integers such that $a^3$ is multiple of $b^2$ and $b-1$ is multiple of $a-1$ .

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samrocksnature

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samrocksnature

#2 h May 17, 2022, 7:32 PM • 1 Y

Y by ImSh95

HELP??!?!

This post has been edited 8 times. Last edited by samrocksnature, May 17, 2022, 10:42 PM

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electrovector

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electrovector

#3 h May 17, 2022, 7:37 PM • 7 Y

Y by ImSh95, Greenhand002, Vlad2004, sabkx, jannatiar, ehuseyinyigit, M.Daniyal

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.

This post has been edited 3 times. Last edited by electrovector, May 17, 2022, 7:50 PM

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RagvaloD

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RagvaloD

#4 h May 17, 2022, 7:39 PM • 4 Y

Y by samrocksnature, ImSh95, Doppel, Mango247

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

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samrocksnature

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samrocksnature

#5 h May 17, 2022, 7:39 PM • 5 Y

Y by ImSh95, Mango247, Mango247, Mango247, channing421

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 $3.$

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motannoir

171 posts

motannoir

#6 h May 17, 2022, 7:46 PM • 2 Y

Y by ImSh95, Greenhand002

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.

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VicKmath7

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VicKmath7

#7 h May 17, 2022, 7:49 PM • 9 Y

Y by Iora, ImSh95, SunnySeattle, Numbertheorydog, Greenhand002, SADAT, rstenetbg, GeorgeRP, AlexCenteno2007

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 $a=b$ .

This post has been edited 5 times. Last edited by VicKmath7, May 17, 2022, 8:24 PM

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grupyorum

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grupyorum

#8 h May 17, 2022, 8:42 PM • 3 Y

Y by ImSh95, Mahmood.sy, Greenhand002

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.

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MathLuis

1524 posts

MathLuis

#9 h May 17, 2022, 9:09 PM • 3 Y

Y by ImSh95, Greenhand002, ACBY

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

This post has been edited 1 time. Last edited by MathLuis, Feb 12, 2023, 12:21 AM

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square_root_of_3

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square_root_of_3

#10 h May 17, 2022, 11:05 PM • 5 Y

Y by ImSh95, Greenhand002, mannshah1211, Jack_w, RecursiveCo

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.

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CANBANKAN

1301 posts

CANBANKAN

#11 h May 18, 2022, 12:06 AM • 2 Y

Y by ImSh95, Greenhand002

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.

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L567

1184 posts

L567

#12 h May 18, 2022, 2:24 AM • 2 Y

Y by ImSh95, Greenhand002

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 $(n,n)$ . $\blacksquare$

This post has been edited 1 time. Last edited by L567, May 21, 2022, 2:49 AM

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Quidditch

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Quidditch

#13 h May 18, 2022, 2:46 AM • 4 Y

Y by ImSh95, Greenhand002, Mai-san, BEN_PHAN

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.
$\boxed{(a,b)=(n,1),(n,n)\text{ for every positive integers }n.}$

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MrOreoJuice

594 posts

MrOreoJuice

#14 h May 18, 2022, 5:53 AM • 2 Y

Y by ImSh95, Greenhand002

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.

This post has been edited 2 times. Last edited by MrOreoJuice, May 23, 2022, 4:06 PM

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sanyalarnab

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sanyalarnab

#15 h May 18, 2022, 7:34 AM • 4 Y

Y by ImSh95, Timmy456, GoodMorning, Greenhand002

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.

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Nguyenhuyhoang

207 posts

Nguyenhuyhoang

#50 h May 30, 2024, 7:10 PM

Y by

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 $a$ .

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Nguyenhuyhoang

207 posts

Nguyenhuyhoang

#51 h May 30, 2024, 7:11 PM

Y by

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 $a$ .

This post has been edited 1 time. Last edited by Nguyenhuyhoang, May 30, 2024, 7:56 PM

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Ywgh1

139 posts

Ywgh1

#53 h Aug 27, 2024, 11:54 AM

Y by

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 $(n,1)$ .

This post has been edited 1 time. Last edited by Ywgh1, Aug 27, 2024, 11:54 AM

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AshAuktober

1005 posts

AshAuktober

#54 h Aug 27, 2024, 2:49 PM

Y by

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 $t \in \mathbb{N}$ . $\square$

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SomeonesPenguin

128 posts

SomeonesPenguin

#55 h Aug 27, 2024, 6:23 PM

Y by

Here is a quick solution. :-D

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. $\square$

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alexanderhamilton124

395 posts

alexanderhamilton124

#56 h Sep 1, 2024, 6:55 PM

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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.

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RedFireTruck

4223 posts

RedFireTruck

#57 h Sep 4, 2024, 11:04 PM

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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 $\boxed{a=b}$ .

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Sadigly

183 posts

Sadigly

#58 h Dec 25, 2024, 1:25 PM • 1 Y

Y by alexanderhamilton124

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

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math004

23 posts

math004

#59 h Feb 2, 2025, 7:20 AM

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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.

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pie854

243 posts

pie854

#60 h Feb 2, 2025, 11:21 AM

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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 $a=b$ .

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SSS_123

20 posts

SSS_123

#61 h Feb 3, 2025, 5:16 PM

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Given conditions,
$b^2|a^3$
$a-1|b-1$
where $a,b$ are positive integers.
Let $a,b \neq 1$
From the 2nd condition we have,
$a-1 \leq b-1$
$a \leq b$
From the first condition we have,
$b^2|a^3$
So we can write $a^3=b^2k$ for some positive integer $k$ .

Now
$b-1 \equiv 0 \mod{(a-1)}$
$b \equiv 1 \mod{(a-1)}$
$b^2 \equiv 1 \mod{(a-1)}$
$b^2k \equiv k \mod{(a-1)}$
$a^3 \equiv k \mod{(a-1)}$
$1 \equiv k \mod{(a-1)}$
$1 \equiv \frac{a^3}{b^2} \mod{(a-1)}$
$\frac{a^3}{b^2} - 1 \equiv 0 \mod{(a-1)}$
So
$a-1|\frac{a^3}{b^2} - 1$
$a-1|\frac{a^3-b^2}{b^2}$
$b^2(a-1)|a^3-b^2$
$ab^2-b^2|a^3-b^2$
$ab^2-b^2 \leq a^3-b^2$
$ab^2 \leq a^3$
$b^2 \leq a^2$
$b \leq a$
But we also got $a \leq b$
Thus $a=b$
Now if $a=1$ then it is obvious that the conditions won't fullfill
if $b=1$ ,it is easy to see that it works for all positive integers $a$
Thus the required solutions are:
$(a,b)=(x,1),(x,x)$ for all positive integers $x$ .

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Iveela

116 posts

Iveela

#62 h Feb 28, 2025, 6:47 PM

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Answer: $(a, b) = (x, 1)$ and $(x, x)$ .

Suppose there exist other solutions. Let $b - 1 = k(a - 1)$ which we can rewrite as $b = (a - 1)k + 1$ . Then $d = \text{gcd}(a, b) = \text{gcd}(a, k - 1) \leq k - 1$ . Now, notice that $b^2 \mid a^3 \Leftrightarrow \left (\frac{b}{d} \right)^2 \mid \left( \frac{a}{d} \right)^3 \cdot d$ implies $\left( \frac{b}{d} \right)^2 \mid d$ . Consequently, we get the following bound for $k$ .
$\frac{b^2}{d^2} \leq d \implies a^2 \leq b^2 \leq d^3 \leq (k - 1)^3 \implies a^{\frac{2}{3}} + 1 \leq k.$ Plugging this back in, we receive $b \geq (a - 1)(a^{\frac{2}{3}} + 1) + 1 \geq a^{\frac{5}{3}}$ which implies $b^2 > a^3$ , a contradiction. $\square$

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ray66

35 posts

ray66

#63 h Mar 12, 2025, 6:18 AM

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$(a,b)=(x,1)$ and $(x,x)$ for a positive integer $x$ .

First, verify that $b=1$ works. Next, assume $b \ge 2$ . Then $b \ge a$ from $b-1$ is a multiple of $a-1$ . Also, $a^3=qb^2$ . If $q=a$ , then $a=b$ . Otherwise, $b^2 \equiv 0 \pmod a$ so $a | b$ . Writing $b-1 = k(a-1)$ gives $b-1 \equiv -k \pmod a$ or $k \equiv 1 \pmod a$ . $k=a+1$ gives $b=a^2$ , but then $a^4 | a^3$ , contradiction. So $k=1$ and $a=b$ .

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Ilikeminecraft

626 posts

Ilikeminecraft

#64 h Apr 25, 2025, 4:09 PM

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By the first condition, we have that $a^3 = kb^2$ for some $k\in\mathbb Z.$ Hence, $a - 1 \mid b - 1 \mid b^2 - 1 \mid kb^2 - k = a^3 - k.$ Hence, we conclude $k\equiv 1\pmod {a - 1}.$ If $k\neq a,$ then $k\geq 2a - 1,$ which is clearly absurd since our second condition implies $b \geq a.$ Thus, $k = a,$ and so $\boxed{a = b > 1}$ is the only solution.

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NerdyNashville

14 posts

NerdyNashville

#65 h Apr 28, 2025, 6:43 AM

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Solution

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