The rest two unknown

36 + 16 5 = a + b c \large \sqrt{36 + 16\sqrt5 }= a + b\sqrt c

If positive integers a a , b b and c c satisfy the equation above, then which of the following value is possible for a b c abc ?

40 11 45 20

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1 solution

36 + 16 5 = a + b c \sqrt{36 + 16\sqrt{5}} = a + b\sqrt{c}

Squaring both sides gives us

( 36 + 16 5 ) 2 = ( a + b c ) 2 (\sqrt{36 + 16\sqrt{5}})^2 = (a + b\sqrt{c})^2

36 + 16 5 = a 2 + b 2 c + 2 a b c 36 + 16\sqrt{5} = a^2 + b^2c + 2ab\sqrt{c}

Therefore,

a 2 + b 2 c = 36 a^2 + b^2c = 36

2 a b = 16 2ab = 16

c = 5 c = 5

Since 2 a b = 16 2ab = 16 , a b = 8 ab = 8 .

Finally, a b c = 40 abc = 40 .

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