Question 251112
LET X, X+2 & X+4 BE THE THREE INTEGERS.
2(X+4)^2=104+X(X+2)
2(X^2+8X+16=104+X^2+2X
2X^2+16X+32=104+X^2+2X
2X^2-X^2+16X-2X-104+32=0
X^2+14X-72=0
(X+18)(X-4)=0
X+18=0 
X=-18 ANS.
X-4=0
X=4 ANS.
PROOF:
2(4+4)^2=104+4(4+2)
2*8^2=104+4*6
2*64=104=24
128=128
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2(-18+4)^2=104-18(-18+2)
2*-14^2=104-18*-16
2*196=104+288
392=392