Find the sum of divisors of 544 which are perfect squares.

• 544 = 2^5 * 17
  
  perfect squares divisiors are 1, 2^2 & 2^4 only
  
  sum of divisors of 544 which are perfect squares = 1 + 2^2 + 2^4 = 21
  
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• sum of divisors 23
  
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• do we consider 1 also ?
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• 544 = 2^5 * 17^1
  perfect square=(2^0,2^2,2^4 & 17^0)
  sum is (2^0+2^2+2^4)(17^0)
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