If the sum of the squares of three consecutive integers is 194, what are the numbers?

#a^2 + b^2 + c^2 = 194#

There are two possible scenarios, either the three numbers go even-odd-even or they go odd-even-odd. Any even number can be written as #2k, kinZZ# and any odd number can be written as #2j+1, jinZ#. For the first scenario, we have:

#(2k)^2 + (2k+1)^2 + (2(k+1))^2 = 194#

#(2k)^2 + (2k+1)^2 + (2k+2)^2 = 194#

#4k^2 + 4k^2 + 4k + 1 + 4k^2 + 8k + 4 = 194#

#12k^2+12k+5 = 194#

We now have a quadratic we can solve for k.

#12k^2 + 12k - 189 = 0#

#k = (-12+-sqrt(144-4(12)(-189)))/24 = 7/2 or -9/2#

These are not integer values so this is not the case.

Trying the second case, we have:

#(2k+1)^2 + (2(k+1))^2 + (2(k+1)+1)^2 = 194#

#(2k+1)^2 + (2k+2)^2 + (2k+3)^2 = 194#

#4k^2+4k+1 + 4k^2+8k+4 + 4k^2+12k+9 = 194#

#12k^2+24k+14=194#

#12k^2+24k-180=0#

#12(k^2+2k-15) = 0#

#12 != 0 implies k^2+2k-15=0#

Factorising our quadratic gives;

#(k+5)(k-3) = 0#

so k is 3 or -5. This means our numbers are:

#7,8,9# or #-9,-8,-7#

Let the three numbers be #x-1#, #x# and #x+1#. As sum of their squares is #194#, we have

#(x-1)^2+x^2+(x+1)^2=194# or

#x^2-2x+1+x^2+x^2+2x+1=194# or

#3x^2+2=194# or #3x^2-192=0#

Or #x^2-64=0# i.e. #(x+8)(x-8)=0#

Hence #x=-8# or #x=8# and as this is middle number

Three consecutive integers are #{-9,-8,-7}# or #{7,8,9}#.