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Question: What is the least number that is a sum of three squares of primes in exactly six ways?

... I know it is not research mathematics. Happy new year!

EDIT: Now that it is answered I should note that I learned this puzzle from a tweet by Ed Southall. I thought it is fun to share.

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closed as off-topic by Wojowu, Ben Linowitz, Ben Barber, Chris Godsil, Neil Hoffman Jan 5 at 1:35

This question appears to be off-topic. The users who voted to close gave these specific reasons:

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    $\begingroup$ If you know it is not research mathematics, why post it here? Mathematics SE exists. $\endgroup$ – Wojowu Jan 4 at 12:25
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    $\begingroup$ this is more like a puzzle, right? the answer is 2019 $$2019=a^2+b^2+c^2,\;\;\text{with}\;\;(a,b,c)\in\{(7,11,43),(7,17,41),(13,13,41),(11,23,37),(17,19,37),(23,23,31)\}$$ $\endgroup$ – Carlo Beenakker Jan 4 at 12:43
  • $\begingroup$ @CarloBeenakker, you are right. $\endgroup$ – Andreas Thom Jan 4 at 12:47
  • $\begingroup$ Is there an elegant or intuitive solution? Of course, one can easily find the answer with a computer program. $\endgroup$ – MathematicsStudent1122 Jan 4 at 17:35
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the answer is 2019

$$2019=a^2+b^2+c^2,\;\;\text{with}\;\;(a,b,c)\in\{(7,11,43),(7,17,41),(13,13,41),(11,23,37),(17,19,37),(23,23,31)\}.$$

I think this was first noticed by Ed Southall

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The sequence $f(n)$ given by those smallest integers which can be written in $n$ ways as the sum of squares of three primes has first eleven values,

$$12,219,363,699,1179,2019,2259,3891,4059,6459,5379.$$

As one can see, the $10$th value, $6459$, is larger than the $11$th value, $5379$. This is OEIS A214512, which indicates that it was really T.D. Noe who observed this earlier.

In any case it will be some time before this puzzle can be used like this again.

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