Let $n$ be a positive integer and there be $p_1,p_2,p_3,........p_n$ prime numbers such that all of them are greater than $5$. If $6$ divides $p_1 ^2 + p_2 ^2 + p_3 ^2+\ldots +p_n ^2$, prove that $6$ divides $n$.
NOTE:- This problem is the $2nd$ question of $1998$ $RMO$ (Regional Mathematical Olympiad).
I tried using congruences with $2$ and $3$ on the first condition but it does not work.
To Show that $6$ divides $n$ if $6$ $|$ $(p_1 ^2 + p_2 ^2 + p_3 ^2 ......... p_n ^2)$ where $p_i$ is a prime$
5
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elementary-number-theory
contest-math
divisibility
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0Do you mean $p_1 ^2 + p_2 ^2 + p_3 ^2 + \cdots + p_n ^2$ ? – 2017-01-07
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0Yes. By "the first condition" ,I mean $6|(p_1 ^2 + p_2 ^2 + p_3 ^2 ......... p_n ^2)$ – 2017-01-07
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0RMO stands for Regional or Romanian Mathematical Olympiad? – 2017-01-07
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0The convention with $\cdots$ in sums is you put a $+$ (or in some cases $-$ or $+-$) between the $\cdots$ and the adjacent terms. If you write $p_3^2\cdots p_n^2$ it usually means you _multiply_ the numbers. To add them, write $p_3^2+\cdots+p_n^2$. Also note that many people put the $\cdots$ in the middle of the line, not at the bottom like $\ldots$, although this detail won't cause confusion. – 2017-01-07
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0@scummy, it is for regional mathematical Olympiads.... (Held in different states of India) – 2017-01-07
3 Answers
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Notice that every prime$>3$ is of the form $6n\pm1$, So square of every such prime will be $36n^2\pm12n+1$, so when you are adding all those squares, you will get something like:
$36(n_1^2+n_2^2+n_3^2......)+12(\pm n_1+\pm n_2.....)+ n(1)$..
The two terms containing $6$ and $12$ are divisible by $6$, the divisibility of whole by $6$ depends upon divisibilty of $n$ by $6$ and vice versa.
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1$36n^2\pm 12n+1$. – 2017-01-07
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0great answer, thanks! – 2017-01-07
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0Welcome @A.Molendijk – 2017-01-07
6
The sum of squares of $n$ odd numbers is $\equiv n\pmod 8$ because a single odd square is $\equiv 1\pmod 8$; similarly the sum of squares of $n$ numbers not divisible by $3$ is $\equiv n\pmod 3$. Hence under the given conditions, we even have $p_1^2+\ldots +p_n^2\equiv n\pmod {24}$, but of course in particular $p_1^2+\ldots +p_n^2\equiv n\pmod {6}$.
3
Using congruences on $2$ and $3$ works as follows:
For all primes $p_i\ge 5$, we know that $2 \nmid p_i$ and $3\nmid p_i$. Therefore $p_i^2\equiv 1 \bmod 2$ and $p_i^2\equiv 1 \bmod 3$, which immediately gives us that $p_i^2 \equiv 1 \bmod 6$ by the Chinese Remainder Theorem.
Clearly adding $n$ such prime squares to a total $s$ will only give us $s\equiv 0\bmod 6 $ (that is, $s\mid 6$) if $n\mid 6 $ also.