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Let $a,b,c,d,e,f,g$ be positive integers greater than or equal to $2$. What values of these numbers satisfy the equations $$a+b+c+d+e+f+g =18 \tag 1$$ $$a(b+c+d+e+f+g+3) + b(c+d+e+f+g+3) + c(d+e+f+g+3) + d(e+f+g+3) + e(f+g+3) + f(g+3) + 3g = 188 \tag 2$$

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Equation (2) is just $\displaystyle \sum_{\text{sym}} ab + 3\sum_{sym} a = 188$. Using (1) we then get $\displaystyle \sum_{\text{sym}} ab = 134$.

Now squaring (1) and subtracting twice the above we get $\displaystyle \sum_{\text{sym}} a^2 = 56$.

Now, at least $3$ of the variables are $2$, because otherwise their sum is greater than $18$. Suppose $e,f,g$ are $2$, then we get $a+b+c+d = 12$ and $a^2+b^2+c^2+d^2 = 44$.

Now $4\cdot 3^2 < 44$, so we must have another variable equal to $2$ (say $d$). Hence we get $a+b+c = 10$ and $a^2+b^2+c^2 = 40$. $(a,b,c) = (3,3,4)$ still has $a^2+b^2+c^2$ too small, so another variable must be $2$, say $c$. So we get $a+b=8$, $a^2+b^2 = 36$. $4^2+4^2$ is too small, as is $3^2 + 5^2$, while $2^2 + 6^2$ is too big.

Thus there are no solutions.

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  • $\begingroup$ Sorry to anyone confused by all the edits. I did a "sanity check" on the last one and made a mistake on my sanity check and thought that I had a mistake in the above. $\endgroup$
    – universalset
    Dec 7, 2013 at 13:57
  • $\begingroup$ What is "sym" under the sum? $\endgroup$
    – Parseval
    Dec 7, 2013 at 14:00
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    $\begingroup$ It means "symmetric". That is, I want the appropriate symmetric polynomial with one of its terms equal to the given term: in the case of $\sum_{\text{sym}} a$ that's just $a+b+c+d+e+f+g$; in the case $\sum_{\text{sym}} ab$ I want to sum over all (unordered) pairs, getting $ab+ac+\cdots + fg$. $\endgroup$
    – universalset
    Dec 7, 2013 at 14:03
  • $\begingroup$ Can you please elaborate on "Now at least 3 of the variables are 2,because otherwise their sum is greater than 18."? $\endgroup$
    – rah4927
    Dec 7, 2013 at 14:21
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    $\begingroup$ $\displaystyle \sum_{\text{sym}} ab$ is just $a(b+c+d+e+f+g)+b(c+d+e+f+g)+c(d+e+f+g)+d(e+f+g)+e(f+g)+fg$.Note that this is just equation(2) with $3a+3b+3c+3d+3e+3f+3g$ missing.But $3a+3b+3c+3d+3e+3f+3g$ is just $3\displaystyle \sum_{\text{sym}} a$. To get equation (2),we need to add this to $\displaystyle \sum_{\text{sym}} ab$. That is why $\displaystyle \sum_{\text{sym}} ab$$+$$3\displaystyle \sum_{\text{sym}} a$ is equal to equation (2). $\endgroup$
    – rah4927
    Dec 8, 2013 at 5:59

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