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Source: ISL 2023 N7

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1748 posts

OronSH

#1 h Jul 17, 2024, 12:00 PM • 15 Y

Y by ehuseyinyigit, peace09, kred9, KevinYang2.71, GeoKing, MarkBcc168, khina, aidan0626, Sedro, Tqhoud, megarnie, ohiorizzler1434, ihatemath123, Yiyj, Jack_w

Let $a,b,c,d$ be positive integers satisfying $\frac{ab}{a+b}+\frac{cd}{c+d}=\frac{(a+b)(c+d)}{a+b+c+d}.$ Determine all possible values of $a+b+c+d$ .

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math90

#2 h Jul 17, 2024, 12:01 PM • 2 Y

Y by OronSH, ihatemath123

Solution

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1748 posts

OronSH

#3 h Jul 17, 2024, 12:07 PM • 7 Y

Y by peace09, KevinYang2.71, HyperDunteR, math90, megarnie, Alex-131, Yiyj

The answer is all non-squarefree positive integers.

For the construction, take $(a,b,c,d)=(m,m(n-1),m(n-1),m(n-1)^2)$ giving $a+b+c+d=mn^2$ for $n\ge 2.$

To prove necessity rearrange it to $ab(c+d)+cd(a+b)=\frac{(a+b)^2(c+d)^2}{a+b+c+d},$ so the RHS is an integer. Thus if $s=a+b+c+d$ works there must exist $x=a+b,y=c+d$ such that $x+y=s$ and $s\mid x^2y^2.$ Clearly $s\ne 1$ by size, now if $p\mid s$ then $p\mid x^2y^2.$ Suppose $p\mid x,$ then $p\mid y$ from $x+y=s.$ Thus any prime dividing $s$ also divides both $x,y.$ But if $s$ is squarefree, it is the smallest positive integer with all of its prime factors. Thus $x,y\ge s$ so $x+y\ge 2s,$ contradiction.

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#4 h Jul 17, 2024, 12:43 PM • 1 Y

Y by HaO-R-Zhe

I will prove that all non-square-free positive integers can be $a+b+c+d$ .

We have $(a+b+c+d)(ab(c+d) + cd(a+b)) = (a+b)^2(c+d)^2$ .

Notice that $n(k+1)^2$ works, because we can choose $a = nk^2$ , $b = kn$ , $c = kn$ and $d = n$ .

Now, suppose $a+b+c+d = N$ , where $N$ is square-free.
Take some $p|N$ , such that $p$ does not divide $c+d$ . Thus, $p|(a+b)^2(c+d)^2 \implies p|(a+b)^2 \implies p|a+b$ . But $p|N \implies p|c+d$ , contradiction.

Note that we can always find such $p$ because if not, then $N | c+d$ , impossible, because $N > c+d$ .

Thus, all non-square-free positive integers are our solutions.

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5610 posts

#5 h Jul 17, 2024, 12:47 PM • 2 Y

Y by KevinYang2.71, peace09

We claim the answer is all positive integers that are not squarefree.

To see these work, we only need to prove it for $a + b + c + d = p^2$ for all primes $p$ because $(a, b, c, d)$ is a solution iff $(xa, xb, xc, xd)$ is for any positive integer $x$ . Consider $a = 1, b = c = p-1$ , and $d = (p-1)^2$ . Clearly $a + b + c + d = p^2$ . Now, we have $\frac{ab}{a+b} = \frac{p-1}{p}, \frac{cd}{c+d} = \frac{(p-1)^2}{p}, \frac{(a+b)(c+d)}{a + b + c + d} = \frac{p(p^2 - p)}{p^2} = p - 1$ , and clearly $\frac{p-1}{p} + \frac{(p-1)^2}{p} = p - 1$ , so this construction works. Now we prove $a + b + c + d$ can't be squarefree. Suppose otherwise.

We have $(ab(c+d) + cd(a+b))(a + b + c + d) = ((a + b)(c+d) )^2$ Thus, for any prime $p$ dividing $a + b + c + d$ , $p$ divides either $a + b$ or $c + d$ , so $p$ must divide both $a + b$ and $c + d$ . However, this implies the product of distinct primes dividing $a + b + c + d$ divide $a + b$ and $c + d$ . This product equals $a + b + c + d$ since $a + b + c + d$ is squarefree, so $a + b + c + d \mid a + b$ , meaning $a + b + c + d \le a + b$ , absurd.

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407 posts

#6 h Jul 17, 2024, 1:28 PM

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The answer is all non-square-free integers. The construction is $(a,b,c,d) = (kn^2, kn, kn, k)$ for any $n, k\geq 1$ , when both sides equal $kn$ and $a+b+c+d=k(n+1)^2$ . To show these are the only answers, we can rewrite the equation as:
$\mathbb{N} \ni abc+abd+acd+bcd=\frac{(a+b)^2(c+d)^2}{a+b+c+d}.$ Now for any prime $p\mid a+b+c+d$ , $p\mid (a+b)^2(c+d)^2\iff p\mid a+b$ and $p\mid c+d$ . Therefore, if $a+b+c+d$ is square-free and $p\mid a+b+c+d$ , then $p^4\mid (a+b)^2(c+d)^2$ , which is impossible as
$(a+b)^2(c+d)^2<(a+b+c+d)^4 \Longrightarrow (a+b+c+d)^4\nmid (a+b)^2(c+d)^2.$

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1831 posts

#7 h Jul 17, 2024, 1:30 PM • 1 Y

Y by megarnie

There exists a solution whenever $p^2\mid a+b+c+d$ for some prime $p$ . Denoting $n = (a+b+c+d)/p^2$ , the construction is $(a,b,c,d) = (n(p-1), n(p-1)^2, n, n(p-1)),$ which clearly works. Now, suppose $a+b+c+d$ is squarefree. Clear fractions:
$(a+b+c+d)(abc + abd + acd + bcd) =\left((a+b)(c+d)\right)^2.$ Note that we have
$0\equiv \left((a+b)(c+d)\right)^2\equiv (a+b)^4\pmod{a+b+c+d}$ $\implies 0\equiv a+b\pmod{a+b+c+d},$ absurd.

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185 posts

#8 h Jul 17, 2024, 1:38 PM

Y by

Easy for its position. The answer is all non-squarefree positive integers.

Let $s = a + b + c + d$ , assume $s$ is square-free. Multiply $s(a + b)(c + d)$ on both sides, we can see $s \mid (a + b)^2(c+d)^2.$ Let $p$ be a prime divisor of $s$ , then either $p \mid a + b$ or $p \mid c + d$ . Both are equivalent, so $p \mid s \implies p \mid a + b$ holds.
Thus $s \mid a + b$ , and this is clearly a contradiction.

For construction, let $a = 1, b = c = p - 1, d = (p-1)^2$ , we can generate $p^2$ . Multiply some integer to $a, b, c, d$ , we can make all non-squarefree positive integers. $\blacksquare$

This post has been edited 1 time. Last edited by pokmui9909, Jul 17, 2024, 1:39 PM

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blueberryfaygo_55

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blueberryfaygo_55

#9 h Jul 17, 2024, 3:44 PM • 1 Y

Y by megarnie

We claim that $a+b+c+d$ can take all non-squarefree values. We first show that $a+b+c+d$ cannot be squarefree, then provide a working construction for all non-squarefree values.

Let $a+b=m$ , $c+d=n$ , $ab=x$ , and $cd=y$ . The given condition is equivalent to $\begin{align*} \dfrac{x}{m} + \dfrac yn &= \dfrac{mn}{m+n}. \end{align*}$ Cross multiplying yields $(xn+ym)(m+n)=m^2n^2,$ which implies that $m+n \mid m^2n^2$ . Consider a prime $p$ that divides $m+n$ . Then, $p \mid m^2$ or $p \mid n^2$ , meaning that $p \mid m$ or $p \mid n$ , but $p \mid m+n$ , so $p$ must divide both $m$ and $n$ for all prime divisors $p$ of $m+n$ . For the sake of contradiction, suppose that $m+n=a+b+c+d$ is squarefree. Letting $m+n = p_1 p_2 \cdots p_k$ $(k \geq 1)$ , since $m$ is divisible by every prime divisor of $m+n$ , we must have $m = C p_1 p_2 \cdots p_k$ for some positive integer $C$ , but $C p_1 p_2 \cdots p_k \geq p_1 p_2 \cdots p_k,$ so $m \geq m+n$ , which is clearly absurd. Thus, $m+n = a+b+c+d$ cannot be squarefree.

Now if $a+b+c+d$ is non-squarefree, let $a+b+c+d = a^2k$ for positive integers $a,k$ where $a > 1$ . Then, we take $\{a,b,c,d\} = \{k, (a-1)k, (a-1)k, (a-1)^2k\}$ for the given condition to hold. Indeed, we can check that the left hand side is $\begin{align*} \dfrac{ab}{a+b} + \dfrac{cd}{c+d} &= \dfrac{k^2(a-1)}{ka} + \dfrac{k^2(a-1)^3}{k(a-1)a} \\ &= \dfrac{k(a-1)}{a} + \dfrac{k(a-1)^2}{a} \\ &= \dfrac{k(a-1) + k(a-1)^2}{a} \\ &= k(a-1). \end{align*}$ The right hand side is $\begin{align*} \dfrac{(a+b)(c+d)}{a+b+c+d} &= \dfrac{ka \cdot k(a-1)a}{ka+k(a-1)a} \\ &= \dfrac{k^2a^2(a-1)}{ka^2} \\ &= k(a-1) \end{align*}$ which is indeed also the left hand side, so we are done. $\blacksquare$

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#10 h Jul 17, 2024, 9:05 PM

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Much easier when one has seen USA TST 2021/1
as the equation here is equivalent to $(a+b+c+d)(abc+abd+acd+bcd) = (a+b)^2(c+d)^2$ .

If $s=a+b+c+d$ works, so does $sm$ by multiplying each of $a$ , $b$ , $c$ , $d$ with $m$ . Now mimicking the USA problem, take $d=1$ and $a=bc$ to get $s=(b+1)(c+1)$ and $bc(b+1)^2(c+1)^2 = b^2(c+1)^4$ . This is satisfied by $b=c$ , with $s=(b+1)^2$ , hence any non-squarefree integer works.

Conversely, if $p$ is a prime dividing $a+b+c+d$ , then $p$ divides the RHS with power at least $2$ (since RHS is a square) and moreover it divides at least one of $a+b$ and $c+d$ , hence both due to dividing $a+b+c+d$ . But if $a+b+c+d = p_1p_2\cdots p_k$ and each $p_i$ divides $a+b$ and $c+d$ , then in fact the whole RHS divides $a+b$ and $c+d$ separately, which is impossible, as they are smaller in size.

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26 posts

Tqhoud

#11 h Jul 17, 2024, 9:11 PM

Y by

we can check easily that

$a=n,b=nm,c=nm,d=nm^2$
is a solution and

$a+b+c+d=n(m+1)^2$
where $n,m\ge 1$

so all non squarefree is a solution

If $a+b+c+d$ is a squarefree and solution for this problem

We can check that

$(a+b+c+d)(ab(c+d)+cd(a+b))=(a+b)^2(c+d)^2$
let $a+b=t$ and $c+d=r$

so we find that

$t+r|(tr)^2$
and because that $t+r$ is a squarefree

$t+r|tr$
but

$t+r=gcd(t+r,tr)=gcd(t,r)|t<t+r$
wrong result and we are done

This post has been edited 2 times. Last edited by Tqhoud, Jul 17, 2024, 9:12 PM
Reason: .

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180 posts

dkedu

#12 h Jul 17, 2024, 11:12 PM

Y by

e claim the answer is all squareful numbers, that is there exists $m\ge 2$ such that $m^2\mid a+ b+ c+d$ . We provide a construction, let $a+b+c+d = m^2k$ , then take $(a,b,c,d) = (k, k(m-1),k(m-1),k(m-1)^2)$ . We will now show these are the only ones.

WLOG $\gcd(a,b,c,d) = 1$ since scaling an solution does not change whether it works or not. Now assume for the sake of contradiction that $\nu_p(a+b+c+d) \le 1$ for all $p$ prime. Now for all prime $p$ such that $\nu_p(a+b+c+d) =1$ . We WLOG that $\nu_p(a) \le 1, \nu_p(a+b) \le 1$ .

If $\nu_p(a+b) = 0$ , then $\nu_p(c+d) = 0$ , but then $\nu_p(LHS) \ge 0 > -1 = \nu_p(RHS)$ which is a contradiction.

$\nu_p(a+b) = 1$ and $\nu_p(c+d) = n \ge 1$ , so if $\nu_p(a) = 1$ , then $\nu_p(b), \nu_p(c), \nu_p(d) \ge 1$ , so we contradict our assumption that the greatest common divisor is $0$ .

So we have $\nu_p(a+b) = 1, \nu_p(a) = 0$ , so $\nu_p(b) = 0$ and we let $\nu_p(c+d) = l\ge 1$ . Then we have
$-1 \le \nu_p(LHS) = \nu_p(RHS) = l$ so we have $\nu_p(c) + \nu_p(d) = l-1$ meaning $\nu_p(c) = \nu_p(d) = \frac{l-1}{2}$ . So let $l = 2k + 1$ .

However this means that for all $p \mid a+b+c+d$ , $p \mid a+b, c+d$ , so if $a+b+c+d$ is squarefree then $a+b+c+d \mid a+b,c+d$ which is clearly a contradiction so we are done.

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#13 h Jul 18, 2024, 2:37 AM

Y by

not a new (or difficult) idea at all, see USA TST 2021/1, ISL 2005 N3, USAMO 2015/5

The answer is $a + b + c + d$ not squarefree.

Proof of sufficiency. Let $a + b + c + d = pq^2$ with $p$ squarefree. Then
$(a, b, c, d) = (p, p(q-1), p(q-1), p(q-1)^2)$ works.

Proof of necessity. Conversely, assume $a + b + c + d$ is squarefree. Then
$\frac{(a+b)^2(c+d)^2}{a+b+c+d} = ab(c+d)+cd(a+b)$ is an integer, so $a + b + c + d \mid (a+b)^2(c+d)^2$ . Thus
$(a+b)^2(c+d)^2 \equiv (a+b)^2(-a-b)^2 \equiv (a+b)^4 \equiv 0 \pmod{a + b + c + d}$ which implies $a + b + c + d \mid a + b$ since $a + b + c + d$ is squarefree, an obvious size contradiction.

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#14 h Jul 18, 2024, 9:06 AM • 1 Y

Y by GreenTea2593

Absolute disgrace of a problem. Does not deserve this spot.

We will prove that the only solutions are $\boxed{\text{all non-squarefree numbers}}$ .

Necessity Assume $a+b+c+d$ is squarefree. We can rearrange the equation and we get $(abc+abc+acd+bcd)(a+b+c+d)=(a+b)^2(c+d)^2$ Now if we let $p \mid a+b+c+d$ . Then see that $p \mid a+b$ and $p \mid c+d$ . And now by comparing $\nu_p$ , we get $p^3 \mid (abc+abc+acd+bcd) \implies (a+b+c+d)^3 \mid (abc+abc+acd+bcd)$ but this is false by Rearrangement inequality.
Construction See that $(a,b,c,d)=\left(n,n(p-1),n(p-1),n(p-1)^2 \right)$ is a solution which gives us $a+b+c+d=np^2$ , that is non-squarefree.

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#15 h Jul 21, 2024, 12:28 AM

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Solution

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MathPerson12321

3801 posts

MathPerson12321

#17 h Jul 21, 2024, 3:46 PM

Y by

OronSH wrote:

The answer is all non-squarefree positive integers.

For the construction, take $(a,b,c,d)=(m,m(n-1),m(n-1),m(n-1)^2)$ giving $a+b+c+d=mn^2$ for $n\ge 2.$

To prove necessity rearrange it to $ab(c+d)+cd(a+b)=\frac{(a+b)^2(c+d)^2}{a+b+c+d},$ so the RHS is an integer. Thus if $s=a+b+c+d$ works there must exist $x=a+b,y=c+d$ such that $x+y=s$ and $s\mid x^2y^2.$ Clearly $s\ne 1$ by size, now if $p\mid s$ then $p\mid x^2y^2.$ Suppose $p\mid x,$ then $p\mid y$ from $x+y=s.$ Thus any prime dividing $s$ also divides both $x,y.$ But if $s$ is squarefree, it is the smallest positive integer with all of its prime factors. Thus $x,y\ge s$ so $x+y\ge 2s,$ contradiction.

yep
its basically just rearrange the terms then nt i used the same sol
first time i actually solved one of these

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Hermione.Potter

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Hermione.Potter

#18 h Jul 21, 2024, 3:56 PM • 4 Y

Y by sami1618, jp62, Assassino9931, yofro

Once you are able to get the correct answer set (by trying small cases etc.), you're done. I do agree that it's way too simple as a N7. Unfortunately, I did not see this during TST and tripped by not trying $a+b+c+d=9$ (even after correctly knowing to skip C4). Here's another interpretation of the problem:

Attachments:

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168 posts

#19 h Jul 21, 2024, 3:56 PM

Y by

OronSH wrote:

Let $a,b,c,d$ be positive integers satisfying $\frac{ab}{a+b}+\frac{cd}{c+d}=\frac{(a+b)(c+d)}{a+b+c+d}.$ Determine all possible values of $a+b+c+d$ .

It's seemed that there are many solutions......
But I'm curious about whether it has geometric solution (like 2001 IMO P6)
I tried to construct some geometric structure, but I failed
Does anyone have some idea?

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jp62

#20 h Aug 11, 2024, 3:21 PM • 13 Y

Y by Hermione.Potter, ihatemath123, OronSH, sami1618, ehuseyinyigit, Seungjun_Lee, khina, sansgankrsngupta, GrantStar, Therealway, gghx, GM_YoanaMl, yofro

Here's the construction using resistors, as a followup to #18 and #19:

https://i.imgur.com/V3Q0MtA.jpeg

The leftmost circuit has a resistance of $\frac{ab}{a+b}+\frac{cd}{c+d}$ and the rightmost circuit has a resistance of $\frac{(a+b)(c+d)}{a+b+c+d}$ .
For the construction, we set the middle circuit to have the same resistance as both the left and the right.
The first equality is attained when $b=c$ for us to be able to swap the resistors.
The second equality is attained when $ad=bc$ , by the Wheatstone bridge.

This gives us a family of solutions that works. By setting $a=1$ we get $d=b^2$ for a total sum of $a+b+c+d=1+b+b+b^2=(1+b)^2$ , and scaling this up gives $k(1+b)^2$ for any $k,b\in\mathbb N$ .
Thus any non-squarefree positive integer can be attained as the value of $a+b+c+d$ .

It remains to do the NT to prove that for distinct primes $p_1$ through $p_r$ , $a+b+c+d=\prod p_i$ doesn't work.
We clear denominators to get $(a+b+c+d)(abc+bcd+cda+dab)=(a+b)^2(c+d)^2.$ Since $p_i$ divides the LHS, it must also divide the RHS. Notice $p_i\mid a+b\iff p_i\mid c+d$ . Therefore, we get $\nu_{p_i}(RHS)\geq 4$ .
From $\nu_{p_i}(a+b+c+d)=1$ we get $\nu_{p_i}(abc+bcd+cda+dab)\geq3$ .

Combining all these inequalities shows that $abc+bcd+cda+dab\geq\prod p_i^3=(a+b+c+d)^3=3abc+3bcd+3cda+3dab+\text{junk}$ which is clearly false.

Storytime and commentary

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85 posts

#22 h Sep 4, 2024, 1:00 AM

Y by

Here are my thought process for this problem.

First, we observe that the term $a+b$ and $c+d$ appear often, so we let $a+b = x$ , $c+d=y$ , $ab =n$ and $cd = m$ .

We get $(x+y)(mx + ny) = x^2y^2$ Let $\gcd(x,y) = k$ and we replace $x$ with $ki$ and $y$ with $kj$ , so the equation turns into $(i+j)(mi+nj) = k^2i^2j^2$ where $\gcd(x,y) =1$

Note : We can’t suppose without the loss of generality that $k =1$ , since it may result in some of $a,b,c,d$ being an irrational.

The equation we get are something like $\alpha \beta \text{ is perfect square}$ So, we went for a classical technique by observing $\gcd(i+j,mi+nj)$

By euclidian algorithm, we will end up with $\gcd(i+j,pi)$ or $\gcd(i+j,pj)$ , which both of them is $1$

Thus, $i+j$ is forced to be a perfect square.
But, we know that both $\gcd(i+j,j^2)$ and $\gcd(i+j,i^2)$ is $1$ .
So, it need to be square of some divisor of $k$ .
In other words, we can say that $i+j = \frac{k^2}{q^2} \text{for some divisor q of k}$ Thus, $a+b+c+d= \frac{\gcd(a+b,c+d)^3}{q^2}$ Note : It is impossible to have $q = k$ , since it will result in $i+j =1$ , meaning one of $i,j$ need to be $0$ .

So, we have $a+b+c+d$ is always non square free.

But, we know that this equality is homogenized, thus if it is possible to have $a+b+c+d = S$ , then it is also possible to have $a+b+c+d = \epsilon S$

Now, we want to know that what prime number $p$ can’t satisfy $a+b+c+d = p^2$

We have $mx + ny = ( \frac{xy}{p} )^2$
Thus, we have $p \mid x \text{or} p \mid y$ But, since we have $x+y = p^2$ , we have $p \mid x,y$ Suppose $x = wp, y = zp$ , we have $w+z = p$ and the equation $mw + nz = w^2z^2p^2$ Consider mod $p$ $mw - nw \equiv m-n \equiv 0 \pmod{p}$ At this point, the condition we get are quite strong, so we are gonna proceed with some trial and error.
Let $a=1, b=p-1, c = p-1, d= p^2-2p +1$ . We are gonna see that this tuplet works. So, the answer is $\boxed{ a+b+c+d \text{ is a non square free integers}}$

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SomeonesPenguin

129 posts

SomeonesPenguin

#23 h Nov 14, 2024, 7:28 PM • 1 Y

Y by zzSpartan

The answer is all non squarefree positive integers.

It suffices to show that all the squares work since if $(a,b,c,d)$ is a solution, then $(ka,kb,kc,kd)$ is as well. For this, just pick $(a,b,c,d)=(1,p,p,p^2)$ .

Note that the condition is equivalent to $(a+b+c+d)(abc+abd+acd+bcd)=(a+b)^2(c+d)^2$ And suppose that $a+b+c+d$ is squarefree and let $a+b=x$ and $c+d=y$ . We have $x+y\mid xy$ since $x+y$ is squarefree and this implies $x+y\mid x^2$ so $x+y\mid x$ , which is a contradiction. $\blacksquare$

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#24 h Dec 16, 2024, 5:07 AM • 1 Y

Y by RainbowJessa

The hardest part of the problem is to not overthink it. :/

The answer is all non-squarefree positive integers.

Construction: It suffices to construct for perfect squares $n^2$ . In this case, take $(a, b, c, d) = (1, n-1, n-1, n(n-1))$ , so
$\frac{ab}{a+b} + \frac{cd}{c+d} = \frac{n-1}n + \frac{(n-1)^2}n = n-1 = \frac{n^2(n-1)}{n^2} = \frac{(a+b)(c+d)}{a+b+c+d}.$ Bound: This proof is so stupid it's funny. We may assume $\gcd(a, b, c, d) = 1$ . Let $p \mid a+b+c+d$ . Because $\frac{(a+b)^2(c+d)^2}{a+b+c+d} = ab(c+d)+cd(a+b)$ is an integer, $p \mid a+b$ and $p \mid c+d$ . On the other hand, if $p \mid a+b$ and $p \mid c+d$ , then clearly $p \mid a+b+c+d$ ; so $a+b+c+d$ consists of the products of the primes $p$ that divide both $a+b$ and $c+d$ . But then $a+b+c+d \mid \gcd(a+b, c+d)$ , which is clearly impossible for size issues.

Remark: The $ab(c+d)+cd(a+b)$ is completely useless, and I spent too much time trying to figure out its significance.

This post has been edited 1 time. Last edited by HamstPan38825, Dec 16, 2024, 5:10 AM

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awesomeming327.

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awesomeming327.

#25 h Apr 27, 2025, 5:57 PM • 1 Y

Y by RainbowJessa

The answer is non-squarefree.

Sufficient: Let $n=x^2y$ where $x>1$ . Then $(y,y(x-1),y(x-1),y(x-1)^2)$ just works.
Necessary: Suppose $a+b+c+d$ is squarefree. If $p\mid a+b+c+d$ then either $a+b$ and $c+d$ are both divisible by $p$ or they both aren't. If they both aren't then $\nu_p$ of the LHS is nonnegative while the $\nu_p$ of the RHS is negative, impossible. Therefore, $p\mid a+b$ . Since $a+b+c+d$ is squarefree, $a+b=a+b+c+d$ , impossible.

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sansgankrsngupta

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sansgankrsngupta

#26 h May 6, 2025, 6:30 AM

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Sunshine132 wrote:

I will prove that all non-square-free positive integers can be $a+b+c+d$ .

We have $(a+b+c+d)(ab(c+d) + cd(a+b)) = (a+b)^2(c+d)^2$ .

Notice that $n(k+1)^2$ works, because we can choose $a = nk^2$ , $b = kn$ , $c = kn$ and $d = n$ .

Now, suppose $a+b+c+d = N$ , where $N$ is square-free.
Take some $p|N$ , such that $p$ does not divide $c+d$ . Thus, $p|(a+b)^2(c+d)^2 \implies p|(a+b)^2 \implies p|a+b$ . But $p|N \implies p|c+d$ , contradiction.

Note that we can always find such $p$ because if not, then $N | c+d$ , impossible, because $N > c+d$ .

Thus, all non-square-free positive integers are our solutions.

OG! WOW

This post has been edited 1 time. Last edited by sansgankrsngupta, May 6, 2025, 6:30 AM
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