All the coefficients of $f(x)=(x+a)(x+b)(x+c)-(x-d)(x-e)(x-f)=$ $Sx^2+(ab+bc+ca-de-ef-fd)x+(abc+def)$ are multiples of $S$. Evaluating $f$ at $d$ we get that $f(d)=(a+d)(b+d)(c+d)$ is a multiple of $S$. So this implies that $S$ is composite, since $a+d,b+d,c+d$ are all strictly less than $S$. davron

My solution (some kind of minimality idea). Suppose $S$ is prime. We observe that if $(a, b, c, d, e ,f)$ is a solution then $(a-1, b-1, c-1, d+1, e+1, f+1)$ is also a solution. Note that $S$ does not change. Denote $X=\min\{a,b,c\}$. We can claim that $(a-X, b-X, c-X, d+X, e+X, f+X)$ also satisfies the conditions. So we get $S \mid (d+X)(e+X)(f+X)$. But this is impossible, since $d+X,e+X, f+X$ are all $<S$.

Well, note $(a+d)(b+d)(c+d)=A+dB+d^2S$ now so we get $S|(a+d)(b+d)(c+d)$ , suppose $S$ is prime then it must be less than one of $a+d,b+d,c+d$ but as $S=a+b+c+d$ so absurd and done.

Assume $S$ is a prime. Then $(x-a)(x-b)(x-c)$ and $(x+d)(x+e)(x+f)$ are the same polynomial, mod $S$, so by Lagrange's theorem $\{a,b,c\} = \{-d,-e,-f\}$. Thus $a+b+c+d+e+f \ge 3S > S$.

For the sake of contradiction, let $S$ be prime. Note $S$ divides $(x-a)(x-b)(x-c) - (x+d)(x+e)(x+f)$. Substitute $x= a,b,c,-d,-e,-f,$ we get $S$ divides $(d+a)(d+b)(d+c) , (e+a)(e+b)(e+c), (f+a)(f+b)(f+c)$ and $S|(a+d)(a+e)(a+f)$ $S|(b+d)(b+e)(b+f)$ $S|(c+d)(c+e)(c+f)$ From the first set of 3 conditions, WLOG, we can let $S|(a+d)$, $S|(b+e)$ and $S|(c+f)$. However, each of $(a+d),(b+e), (c+f)$ are smaller than $S$, a contradiction.

Let $p$ be a fixed prime. We say a $6$-tuple $(a,b,c,d,e,f)$ of integers is good if $a+b+c+d+e+f = p$ and $p \mid abc+def$ and $p \mid ab+bc+ca-de-ef-fd$. Claim: If $(a,b,c,d,e,f)$ is good then so is $(a+1, b+1, c+1, d-1, e-1, f-1)$. Proof. Direct computation. $\blacksquare$ We claim there exists no good $6$-tuple of positive integers. If not, WLOG $f$ is the smallest of the six numbers. Then by repeating the claim, the tuple $(a+f, b+f, c+f, d-f, e-f, 0)$ is good. But now $p \mid (a+f)(b+f)(c+f)$ and $p > \max(a+f, b+f, c+f)$, contradiction.

We see that \begin{align*} a+b+c&\equiv -(d+e+f)\pmod{S} \\ ab+bc+ca&\equiv -(de+ef+fd)\pmod{S} \\ abc &\equiv -def\pmod{S}. \end{align*}Now, assuming that $S$ is prime, we see that \[(X-a)(X-b)(X-c)=(X+d)(X+e)(X+f)\]over $\mathbb{F}_S[X]$, so because of unique factorization, we have WLOG that $a\equiv -d\pmod{S}$, $b\equiv -e\pmod{S}$, and $c\equiv -f\pmod{S}$. In particular, this means that $S\mid a+d$. However, $S>a+d$, so this isn't possible. Therefore, $S$ couldn't have been prime to start with. $\blacksquare$

AFSOC $S = p$ for some prime $p$. Define \begin{align*} P(x) &= (x + a)(x + b)(x + c) = x^3 + x^2(a + b + c) + x(ab + bc + ca) + abc \\ Q(x) &= (x - d)(x - e)(x - f) = x^3 - x^2(d + e + f) + x(de + ef + fd) - def, \end{align*}and consider \begin{align*} R(x) &= P(x) - Q(x) \\ &= x^2 \cdot p + x(ab + bc + ca - de - ef - fd) + abc + def. \end{align*}Taking mod $p$, we find that $P(x) \equiv Q(x) \pmod p$. However, plug in $x = d$. We know $Q$ has a root at $d$, so \begin{align*} P(d) = (d + a)(d + b)(d + c) \equiv 0\pmod p, \end{align*}but $p > d + a, d + b, d + c$ and so we are done.

$S\geq6$, so assume for the sake of contradiction that $S$ is prime. Let $P(x)\coloneqq (x-a)(x-b)(x-c)$ and $Q(x)\coloneqq (x+a)(x+b)(x+c)$. Multiplying out, we get that $P(x)\equiv Q(x)\;(\text{mod}\,S)$ for all integers $x$. In particular, we get that for an integer $x$, $S$ divides $P(x)$ if and only if $S$ divides $Q(x)$, which, because $S$ is assumed to be a prime, means that $\{a, b, c\}$ is just some permutation of $\{-d, -e, -f\}$ modulo $S$. WLOG, assume that $a\equiv -d$, $b\equiv -e$ and $c\equiv -f$ modulo $S$. But this means that $a+d$, $b+e$ and $c+f$ are all multiples of $S$, which is impossible, because they are positive integers which sum up to $S$.

Note that $(x+a)(x+b)(x+c)-(x-d)(x-e)(x-f)$ has all of its coefficients divisible by $S$. This implies $S|(d+a)(d+b)(d+c)$. FTSOC assume $S$ is prime. Then it must divide either $d+a,d+b,d+c$, all of which are smaller than $S$, a contradiction, so we are done.

Assume for the sake of contradiction that $S$ is prime. Notice that $$S \mid (a+k)(b+k)(c+k)+(d-k)(e-k)(f-k)$$for all $k \in \mathbb{Z}$. Then taking $k$ to be $-a,-b,-c$ we have that $\{d,e,f\} = \{-a,-b,-c\}$ in $\mathbb{F}_S$ using that $S$ is prime, yet, adding any such values of $a,b,c,d,e,f$, we have that $$S = a+b+c+d+e+f \geq 3S$$which is a contradiction. $\blacksquare$

Suppose that $S$ is a prime and let $x=-d,y=-e,z=-f$,so that $abc \equiv xyz \pmod S$ Now consider the polynomial $ (a+k)(b+k)(c+k)+(d-k)(e-k)(f-k)$ and clearly inputting $x,y,z$ implies $S|(a+d)(a+e)(a+f)$,which implies that $S$ divides one of $(a+d),(a+e),(a+f)$,a contradiction.

Note that the polynomial $f(x) = (x-a)(x-b)(x-c)-(x+d)(x+e)(x+f) = (ab+ac+bc-de-ef-df)x-(abc+def)$ is divisible by $S$ for all integers $x$. Thus, \[S\mid f(a) = -(a+d)(a+e)(a+f)\]To finish, $S$ cannot be prime because $a+d,a+e,a+f<a+b+c+d+e+f = S$, so we're done. $\blacksquare$.

Notice \begin{align*}S\mid f(x)&=Sx^2+(ab+bc+ca-de-df-fd)x+(abc+def)\\&=(x+a)(x+b)(x+c)-(x-d)(x-e)(x-f)\end{align*}Letting $x=d,$ we have $S\mid (d+a)(d+b)(d+c),$ which is absurd if $S$ is prime as $S>d+a,d+b,d+c.$ $\square$

Assume for the sake of contradiction $S$ is prime, and take polynomials $P(x)=(x-a)(x-b)(x-c)$ and $Q(x)=(x+d)(x+e)(x+f).$ Note that \[P(x)\equiv x^3 - (a+b+c)x^2 + (ab+bc+ca)x-abc \pmod S\]Note that on the other hand, \[Q(x)\equiv x^3 + (d+e+f)x^2 + (de+ef+fd)x+def \pmod S\]Now, $Q(x)-P(x)=Sx^2+(de+ef+fd-ab-bc-ca)x+abc+def.$ Note that this is divisible by $S.$ In particular, $Q(a)-P(a)=(a+d)(a+e)(a+f)$ is divisible by $S.$ Note that each of those factors are less than $S$ but $S$ is prime, which is a contradiction.

Observe that we have the system of congruences \begin{align*} a+b+c &\equiv -d-e-f \pmod S \\ ab+bc+ca &\equiv de+ef+fd \pmod S \\ abc &\equiv -def \pmod S. \end{align*}This means that $$(x-a)(x-b)(x-c) \equiv (x+d)(x+e)(x+f) \pmod S$$for all $x \in \mathbb Z$ as the resulting polynomials are equivalent under modulo $S$. Letting $x=a$, $$S \mid (a+d)(a+e)(a+f),$$but $a+d, a+e, a+f < S$, and thus $S$ must be composite.

Suppose otherwise that $S$ is prime $p$. Then we have the following equations. \begin{align*} a+b+c \equiv -(d+e+f) \pmod p \\ ab+bc+ca \equiv de+ef+df \pmod p \\ abc \equiv -def \pmod p \end{align*}We will construct numbers $a_1,b_1, c_1, d_1, e_1, f_1$ such that: \begin{align*} a_1+b_1+c_1 \equiv -(d_1+e_1+f_1) \pmod p \\ a_1b_1+b_1c_1+c_1a_1 \equiv d_1e_1+e_1f_1+d_1f_1 \pmod p \\ a_1b_1c_1 \equiv -d_1e_1f_1 \pmod p \end{align*}In fact, this is not so hard to do. It is just enough to take $a_1=a+1, b_1=b+1, c_1 =c_1$ and $d_1 =d-1, e_1 =e-1, f_1 =f-1$. It is easy to see that they satisfy the first congruence. Note that: \begin{align*} (a+1)(b+1)+(b+1)(c+1)+(c+1)(a+1) \equiv (d-1)(e-1)+(e-1)(f-1)+(f-1)(d-1) \pmod p\\ ab+bc+ca +2(a+b+c)+3 \equiv de+ef+fd -2(d+e+f)+3 \pmod p \end{align*}We see that the second desired congruence is also satisfied by given congruences. Also, observe that: \begin{align*} (a+1)(b+1)(c+1) \equiv -(d-1)(e-1)(f-1) \pmod p \\ a+b+c+(ab+bc+ca)+abc+1 \equiv -((d+e+f)-(de+ef+fd)+def-1) \pmod p \\ (a+b+c+de+f) +(ab+bc+ca-de-ef-fd)+(abc+def) \equiv 0 \pmod p \end{align*}We see that the third desired congruence is also satisfied by given congruences. Since $a,b,c, d,e,f$ are positive integers and their sum it $p$, then they all are less than $p$. WLOG $a$ is the largest among $a,b,c$. With each construction we increase $a,b,c$ by $1$ and decrease $d,e,f$ by $1$. We can keep doing it until $a$ becomes $p$. Suppose that we needed for that $k$ constructions, in other words $a+k =p=a_k$. Consider congruence: $$ a_kb_kc_k \equiv d_ke_kf_k \equiv 0 \pmod p $$We see that it implies that $d_ke_kf_k \equiv 0 \pmod p$. WLOG assume that $d_k = d-k$ is the one which is divisible by $p$. Then $0 \equiv a_k + d_k = a+k+d-k = a+ d \pmod p$, which is contradiction, since it forces $b,c,e,f$ to be $0$. Our initial assumption was wrong, therefore $S$ composite as desired.

I'm not smart as others to see that you can really form a polynomial... Anyways here is a solution by induction which kind of makes the polynomial in a different way which I feel is intuitive too... Solution. Assume on the contrary that $S$ is prime. We will make use of these handy identities in the solution. \begin{align} (a+1)(b+1)(c+1) &= abc + (ab+bc+ca) + (a+b+c) + 1 \\ (d-k)(e-k)(f-k) &= def - (de+ef+df) + (d+e+f) - 1 \end{align}The following claim is the crux of the problem. Claim: $S \mid (a+x)(b+x)(c+x) + (d-x)(e-x)(f-x)$ for all $x \in \mathbb{Z}_{\ge 0}$. Proof: The proof is by induction with the base case being trivial. As an induction hypothesis, assume that $S \mid (a+k)(b+k)(c+k) + (d-k)(e-k)(f-k)$. For ease of notation label $a+k = A$, $d - k = D$ and so on. Now finally for the inductive step, we will just work backward. \begin{align*} S \mid (A+1)(B+1)(C+1) &+ (D-1)(E-1)(F-1) \\ \iff S \mid (ABC + DEF) + (AB+BC+CA) &- (DE+EF+FA) + (A+B+C+\ldots) \end{align*}By the induction hypothesis, we will cancel of $ABC + DEF$ term. It is not hard to see that $S \mid (A+B+C+D+E+F)$ as well. \begin{align*} \iff S \mid AB + BC + CA - DE - EF - FA \end{align*}It isn't hard to expand this out and realize its a tautology. So, the induction is complete.$\square$ Finally choose $x = d$. This would give \[S = a + b + c + d + e + f \mid (a + d)(b + d)(c+d).\]Since $S$ is a prime, it must divide one of the factors. But since all the factors are positive and less than $S$, its the desired contradiction. $\blacksquare$

Let's allow $0$ as value of $d,e,f$. FTSOC, let's assume that $S = p$ is a prime. Here, \begin{align*} S &= a+b+c+d+e+f\\ &= (a+1) + (b+1)+(c+1) + (d-1)+(e-1)+(f-1) \end{align*} Now, $p\mid a+b+c+d+e+f$, $p\mid abc+def$ and $p\mid ab+bc+ca-de-ef-df$. Summing these up gives, $$p\mid (a+1)(b+1)(c+1) + (d-1)(e-1)(f-1)$$Also, \begin{align*} &\sum_{cyc}(a+1)(b+1) - \sum_{cyc}(d-1)(e-1)\\ =& ab+bc+ca-de-ef-df +2(a+b+c+d+e+f) \end{align*} Therefore, $\displaystyle p\mid \sum_{cyc}(a+1)(b+1) - \sum_{cyc}(d-1)(e-1)$. So, if $(a,b,c,d,e,f)$ is a solution then, $(a+1, b+1, c+1, d-1,e-1,f-1)$ is also a solution with $S= p$. Now we continue this process until one of the elements becomes $0$. WLOG that be $f$. Then we will get, $p\mid abc$. But therefore $p$ must divide one of $a,b,c$ which isn't possible as, $p = a+b+c+d+e$. This gives a contradiction. $\blacksquare$

Suppose that $S$ was prime. Then the polynomials $(x+a)(x+b)(x+c)$ and $(x-d)(x-e)(x-f)$ have equivalent expansions modulo $p$, hence we must have $\{-a,-b,-c\} \equiv \{d,e,f\} \pmod{S} \implies \{S-a,S-b,S-c\}=\{d,e,f\}$. But then $d+e+f=3S-a-b-c \implies S=a+b+c+d+e+f=3S$: contradiction. $\blacksquare$

Consider $P(x)=(x+a)(x+b)(x+c)-(x-d)(x-e)(x-f)$ $P(x)=Sx^2+x(ab+bc+ca-ed-ef-df)+(abc+def) \implies S|P(x)$ $\forall x \in \mathbb{Z}$ $P(d)=(a+d)(b+d)(c+d)$ hence as $S|P(d)$. Now , $\mathrm{FTSOC}$ assume $S$ is prime, then as we have $S|(a+d)(b+d)(c+d)$ we have at least one of $(a+d), (b+d) , (c+d)> S$ but since $S=a+b+c+d+e+f$ this is absurd for $a,b,c,d,e, f \in \mathbb{Z}^{+}$ , hence our assumption false and hence $S$ is composite $\blacksquare$.

Solved with xor, xook, and xonk, also known as fuzimiao2013, sixoneeight, popop614. We can rewrite the condition as \[ (t-a)(t-b)(t-c) \equiv (t+d)(t+e)(t+f) \pmod{S} \]for all integer $t$. Suppose for the sake of contradiction that $S$ was prime. Take $t=a$. Then, $S$ divides one of $a+d$, $a+e$, $a+f$. However, since $a,b,c,d,e,f$ are positive integers, that would mean that one of $a+d, a+e, a+f$ is equal to $S$. This is a contradiction as the rest of the variables could not be positive, so $S$ must be composite.

FTSOC let $S$ be a prime $a+b+c \equiv -(d+e+f)$ $abc \equiv (-d)(-e)(-f)$ $ab+bc+ca \equiv (-d)(-e)+(-e)(-f)+(-d)(-f)$ In $\mathbb{F}_S$ $(x-a)(x-b)(x-c) \equiv (x+d)(x+e)(x+f)$ therefore $\{a,b,c\} \equiv \{-d,-e,-f\}$ in $\mathbb{F}_S$ $\therefore$ we're done.

Thanks to torch for suggesting this problem to me! Let $k$ be an integer. Note that $S$ divides the quantity $$abc + def + k(ab+bc+ca-de-ef-fd) + k^2(a+b+c+d+e+f) + 1 - 1 = (a+k)(b+k)(c+k) - (d-k)(e-k)(f-k).$$Now there are $2$ cases. Case 1: The smallest of $a,b,c,d,e,f$ is one of $a,b,c.$ WLOG suppose it is $a.$ Letting $k = -a$ gives $S \mid -(d+a)(e+a)(f+a),$ so switching the positive sign yields $$S \mid (d+a)(e+a)(f+a).$$If $S$ was prime, then it would have to divide at least one of these factors, say $d+a$ (the other cases are symmetrical to this). Then $a+b+c+d+e+f \mid d+a,$ which is a contradiction due to size issues. Case 2: The smallest of $a,b,c,d,e,f$ is one of $d,e,f.$ WLOG suppose it is $d.$ Letting $k = d$ gives $S \mid (a+d)(b+d)(c+d),$ and repeat the size argument from the previous case. Therefore, in any case, $S$ must be composite, as desired.

Used a whole bunch of ARCH Hints after 30 min of failing: FTSOC $S=p$ for some prime $p.$ Let $d_0=-d, e_0=-e, f_0=-f.$ Then we get a vieta-like relation modulo $P.$ This implies that $a,b,c$ and $d_0,e_0,f_0$ are roots to the same polynomial with coefficients modulo $p.$ Now WLOG $a-d_0 \equiv 0\pmod{p},b-e_0 \equiv 0\pmod{p}, c-f_0 \equiv 0\pmod{p}.$ Therefore, $a+d, b+e, c+f$ are multiples of $p$ Therefore, $S=a+d+b+e+c+f=p(k)$ for some integer $k>1$ which violates the condition that $S=p,$ or $S$ is prime. Therefore $S$ must be composite.

Really cute problem. I'm surprised I found the nice solution! We consider ordered pairs of sets $(\{a,b,c\},\{d,e,f\})$ which satisfy the given conditions. We call such a pair a good pair and if $a+b+c+d+e+f=k$ then we say that it is a $k-$good pair. We start off by proving the following nice result about $k-$good pairs. Claim : For all integers $k$ for which there exists a $k-$good pair $(\{a,b,c\},\{d,e,f\})$, the pairs $(\{a-1,b-1,c-1\},\{d+1,e+1,f+1\})$ and $(\{a+1,b+1,c+1\},\{d-1,e-1,f-1\})$ are also $k-$good. Proof : It is not hard to see that they must be $k-$good if they are good at all, since the sum $a+b+c+d+e+f$ does not change in either instance. We show that if $(\{a,b,c\},\{d,e,f\})$ is good then the pair $(\{a-1,b-1,c-1\},\{d+1,e+1,f+1\})$ must also be good. This is a mere calculation. Note that, \[ (a-1)(b-1)(c-1)+(d+1)(e+1)(f+1) = (abc+def) -(ab+bc+ca-de-ef-fd) + (a+b+c+d+e+f)\]where each of the bracketed terms are known to be multiples of $a+b+c+d+e+f$. The proof of the other is entirely similar, so this finishes the proof of the claim. We wish to show that there exists no $p-$good pairs for any prime $p$. By way of contradiction, say there exists such a prime $p$ and a $p-$good pair $(\{a,b,c\},\{d,e,f\})$. Then, WLOG say the minimum element of the set $\{a,b,c,d,e,f\}$ is $a$. By way of our claim, it follows that the pair $(\{0,b-a,c-a\},\{d+a,e+a,f+a\})$ is also $p-$good. Note that here, $d'=d+a$ , $e'=e+a$ and $f'=f+a$ are all positive integers. Then, \[p|(0)(b-a)(c-a)+(d+a)(e+a)(f+a)=d'e'f'\]Since $p$ is a prime, this implies that $p$ divides one of $d'$ , $e'$ and $f'$. But, since $p=0 + (b-a) + (c-a) + (d+a) + (e+a) + (f+a)$ where the first 3 terms are non-negative and the last two terms are strictly positive, \[p = (b-a) + (c-a) + (d+a) + (e+a) + (f+a) \ge d'+e'+f' > d',e',f' \]which is a clear contradiction. Thus, there cannot exist such a prime $p$ which proves the desired result.

Same solution as everyone else.

a bit silly, statement reminded me of usa tst 2021/1 Suppose FTSOC $a+b+c+d+e+f=p$. Claim: For any integer $t$, $$(a+t)(b+t)(c+t)+(d-t)(e-t)(f-t)\equiv 0\pmod p.$$ We have $$(abc+def)+t(ab+ac+bc-de-df-ef)+t^2(a+b+c+d+e+f)+(t^3-t^3)\equiv 0\pmod{p}.$$ In particular, if $t=d$, then $p\mid (a+d)(b+d)(c+d)$. However, $a+d,b+d,c+d$ are all less than $p$, contradiction if $p$ is prime.

Suppose that \(S\) is prime. Observe that \[ a + b + c \equiv -d - e - f \pmod{S} \]\[ abc \equiv -def \pmod{S} \]\[ ab + bc + ca \equiv de + ef + fd \pmod{S} \]By Vieta, we have that for every positive integer \(n\), the following relation holds: \[ (n + a)(n + b)(n + c) \equiv (n - d)(n - e)(n - f) \pmod{S}. \]By setting \(n = -a\), we obtain that \(S\) divides one of the numbers \(-a-d\), \(-a-e\), or \(-a-f\). By symmetry, we can assume without loss of generality that \(S \mid -a-d\). In particular, we have \(-d \equiv a \pmod{S}\), i.e., \(S \mid d+a\). However, this is a contradiction, since \[ a + d < a + b + c + d + e + f = S. \]

Note that by Vieta, we have $S \mid P(x) = (x-a)(x-b)(x-c) - (x+d)(x+e)(x+f)$. But putting $x = a$ we get $S$ cannot be prime, qed.

Assume for the sake of contradiction that $a+b+c+d+e+f$ is prime. Then, $a+b+c \equiv -d-e-f,$ $ab+bc+ca \equiv de+df+ef,$ and $abc \equiv -def,$ when taken modulo $a+b+c+d+e+f.$ Let $i_{1} \equiv -i,$ for $i \in \{a,b,c\}.$ It is clear that $a_{1}+b_{1}+c_{1} \equiv d+e+f,$ and $a_{1}b_{1}+a_{1}c_{1}+b_{1}c_{1} \equiv de+ef+df,$ and $a_{1}b_{1}c_{1} \equiv de+ef+df.$ So, $\{a_{1}, a_{2}, a_{3}\}, \{d, e, f\}$ are the solutions of some cubic modulo $p.$ But, by Lagrange's Theorem, this can only have at most $3$ solutions, so these sets are the same, in some order. So, $i+j \equiv 0,$ where $i \in \{a,b,c\},$ and $j \in \{d,e,f\}.$ But, then $i+j < a+b+c+d+e+f,$ so we have a contradiction.

v_Enhance wrote: Let $p$ be a fixed prime. We say a $6$-tuple $(a,b,c,d,e,f)$ of integers is good if $a+b+c+d+e+f = p$ and $p \mid abc+def$ and $p \mid ab+bc+ca-de-ef-fd$. Claim: If $(a,b,c,d,e,f)$ is good then so is $(a+1, b+1, c+1, d-1, e-1, f-1)$. Proof. Direct computation. $\blacksquare$ We claim there exists no good $6$-tuple of positive integers. If not, WLOG $f$ is the smallest of the six numbers. Then by repeating the claim, the tuple $(a+f, b+f, c+f, d-f, e-f, 0)$ is good. But now $p \mid (a+f)(b+f)(c+f)$ and $p > \max(a+f, b+f, c+f)$, contradiction. When i first saw this problem on math class in 15 minutes i had exactly same solution as you

Note that this implies that the polynomials $P(x) = (x + d)(x + e)(x + f), Q(x) = (x - a)(x - b)(x - c)$ are congruent modulo $S.$ Thus, $P(x) - Q(x) \equiv 0\pmod S.$ However, $P(a) - Q(a) = (a + d)(a + e)(a + f) \equiv 0 \pmod S.$ However, each of the factors are less than $S,$ which finishes.

Suppose, FTSOC, it is a prime. Let $a + b + c + d + e + f = p$. We have $p \mid ab + bc + ca - de - ef - df \implies p \mid abc + c^2(a + b) - dec - efc - dfc \implies p \mid - def + c^2(-c - d - e - f) - dec - efc - dfc \implies p \mid c^3 + c^2d + c^2e + c^2f + dec + efc + dfc + def = (c + d)(c + e)(c + f)$. So, $p$ must divide one of them but all three are less than $a + b + c + d + e + f$, a contradiction. So, $S$ is composite.