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 $\boxed{ a+b+c+d \text{ is a non square free integers}}$

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 $a+b+c+d$.

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.

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.

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.

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 \[(a+b)^2(c+d)^2<(a+b+c+d)^4 \Longrightarrow (a+b+c+d)^4\nmid (a+b)^2(c+d)^2.\]

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.

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. $\blacksquare$

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. $\blacksquare$

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.

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

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.

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.

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.

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

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:

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?

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

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 $\boxed{ a+b+c+d \text{ is a non square free integers}}$

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. $\blacksquare$

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.