can be written as a2+b3+c5, where a,b, and c are positive integers. Find a⋅b⋅c.
Solution:
Expand (a2+b3+c5)2 to obtain 2a2+3b2+5c2+2ab6+2ac10+2bc15, and conclude that 2a2+3b2+5c2=2006,2ab=104,2ac=468, and 2bc=144. Therefore ab=52=22⋅13,ac=234=2⋅32⋅13, and bc=72=23⋅32. Then a2b2c2=ab⋅ac⋅bc=26⋅34⋅132, and abc=23⋅32⋅13=936.
Note that a=bcabc=23⋅3223⋅32⋅13=13 and similarly that b=4 and c=18. These values yield 2a2+3b2+5c2=2006, as required.