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Let p, q, and r be prime numbers such that $$2pqr + p + q + r = 2020$$ Find $$ pq +qr+rp $$ I believe I got the correct answer of $p,q,r$ being $2, 17, 29$ which results in an answer of $585$. I solved it because I knew one of the prime numbers had to be $2$ as $2pqr$ is always even and $p + q + r$ will be odd without $2$ as $2$ is the only even prime where the sum must be even as $2020$ is even. Then setting $p=2$ I was able to see $17$ and $29$ would work. As this was during a math competition I did not use rigorous techniques as I needed to do it quickly.

What could be a fast rigorous way to do this question? Any help would be greatly appreciated, thanks.

healynr
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Il Leone
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    Deduction that one of the terms has to be $2$ is perfectly fine. To find $q,r$, we know we have a term $4qr+q+r$ and some constant. For problems like this, a good idea is trying to factor this thing, possibly after multiplying or adding some constants. In this case, try multiplying by $4$ and adding $1$. – Wojowu May 07 '20 at 15:58
  • Seems like this involves the symmetric polynomials of $p,w,r$. I am certain I have seen an inequality involving the sum of the cubes of the numbers and these expressions – Μάρκος Καραμέρης May 07 '20 at 15:58
  • @Wojowu That should factor to (4r+1)(4q+1)=8073 where factors of 8073 are (1,3,9,13,23,27,39,69,117,207,299,351,621,897,2691,8073) where now just checking to see if one of the factors from (1,69) as sqrt(8073) = 89.84... and 69 works so its pair factor 117 works as well. Is that fastest though? – Il Leone May 07 '20 at 16:05
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    You can use $4q+1\ge 13$ and $4q+1\equiv 1\pmod 4$. – mathlove May 07 '20 at 16:22
  • @mathlove Yes I believe you only have to check 2 values of q and from there find the value of r. – Il Leone May 07 '20 at 16:40
  • Thank you, everyone, for the help! – Il Leone May 07 '20 at 16:41

2 Answers2

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HINT.-A way could be, using the identity $(p+q)^2-(p-q)^2=4pq$, to go from $4pq+(p+q)=2018$, to $$(S+D+\dfrac12)(S-D+\dfrac12)=2018$$ where $S=p+q$ and $D=p-q$ so one has $$(2S+2D+1)(2S-2D+1)=8073=3^3\cdot13\cdot23$$ attention enough gives $$2S+2D+1=3^2\cdot13\Rightarrow S+D=58\\2S-2D+1=3\cdot23\Rightarrow S-D=34$$ which gives $p+q$ and $p-q$.

Piquito
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After you render, let us say, $r=2$, you have

$4pq+p+q=2018$

If we multiply by $4$ and add $1$, the left side factors:

$16pq+4p+4q+1=(4p+1)(4q+1)=8073$

Assuming $p\ge q$ we must have $4q+1\le\sqrt{8073}<90$ so the only choices for $q$ are $2,3,5,7,11,13,17,19$. Since $8073=3^3×13×23$ the available $4q+1$ factors $<90$ are as follows. Red numbers fail to be prime values for $p,q$ and are no solutions. Blue numbers are good solutions.

$4q+1=3^2=9,q=2,4p+1=897,\color{red}{p=224}$

$4q+1=13,q=3,4p+1=621,\color{red}{p=155}$

$\color{blue}{4q+1=69,q=17,4p+1=117,p=29}$

Oscar Lanzi
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