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No Square Sums

Stage: 3 and 4 Short Challenge Level: Challenge Level:2 Challenge Level:2

The following seven pairs add to $16$ so at least one of each pair must be removed:
$(1,15)$, $(2,14)$, $(3,13)$, $(4,12)$, $(5,11)$, $(6,10)$, $(7,9)$.
 
If removing these seven is sufficient, then we would be left with $8$, $16$ and seven others.
 
But
  $16+9 = 25$      So we must remove $9$    and keep its partner $7$
    $7+2 = 9$        So we must remove $2$     and keep $14$
$14+11 = 25$      So we must remove $11$   and keep $5$
    $5+4 = 9$        So we must remove $4$     and keep $12$
$12+13 = 25$      So we must remove $13$   and keep $3$
    $3+1 = 4$        So we must remove $1$     and keep $15$
$15+10 = 25$      So we must remove $10$   and keep $6$
 
But we have kept $3$ and $6$ which add to $9$.
 
Hence it is not sufficient to remove only seven. If we remove the number $6$, we obtain a set which satisfies the condition: $\{8,16,7,14,5,12,3,15\}$ or in ascending order $\{3,5,7,8,12,14,15,16\}$.
 
Hence eight is the smallest number of numbers that may be removed.

 
 

This problem is taken from the UKMT Mathematical Challenges.
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