If with , and positive integers, what digit will always be in the product ?
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Start with a + b + c 3 = 1 + 3 . Square both sides: a + b + c 3 = 4 + 2 3 . Subtract a : b + c 3 = ( 4 − a ) + 2 3 . Square both sides: b + c 3 = ( 4 − a ) 2 + 1 2 + 4 ( 4 − a ) 3
Now, if c = 4 ( 4 − a ) , then 3 = c − 4 ( 4 − a ) ( 4 − a ) 2 + 1 2 − b would be rational. Since it's known to be irrational, we must have both c = 4 ( 4 − a ) , b = ( 4 − a ) 2 + 1 2 Further, since we have assumed a , c are positive integers, we must have a ∈ { 1 , 2 , 3 } . We now consider each case:
We can see that the only digit appearing in all three in base 10 is 5 .
Sidenote: When you square both sides of an equation, you can introduce extraneous solutions only if one side is negative prior to squaring. Therefore, to show each of these is a valid solution to the initial equation, it's enough to note that for 1 ≤ a ≤ 3 , both sides of the given equation are positive prior to squaring both sides.