As Long As 128 Divides My Square

Find the sum of all solutions to

n 2 0 ( m o d 128 ) , n^2 \equiv 0 \pmod {128},

where n n is an integer satisfying 0 n < 128 0 \leq n < 128 .


The answer is 448.

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1 solution

Alex G
May 3, 2016

Relevant wiki: Modular Arithmetic - Problem Solving - Basic

For a number to be 0 m o d 128 0 \mod 128 , 128 128 must divide that number. For 128 128 to divide a number, the number must have 2 7 2^7 in its prime factorization. Note that squaring a number does not change its prime factorization, it simply doubles the powers of the primes. Therefore, n n must include 2 4 2^4 in its prime factorization. Other solutions will simply be multiples of 2 4 2^4 , as 128 128 has no primes other than 2 2 in its prime factorization. Taking the sum of all multiples of 2 4 2^4 that are < 128 <128 :

2 4 ( 1 + 2 + 3 + 4 + 5 + 6 + 7 ) 2^4(1+2+3+4+5+6+7)

2 4 ( 28 ) 2^4(28)

448 \boxed{448}

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