Integers a and b are such that #(a + 3sqrt(5))^2# + #a - b sqrt(5)# = 51. Find the possible values of a and corresponding values of b....?

1 Answer
Apr 4, 2017

#a = -3# and #b=-18# or #a=2# and #b=12#

Explanation:

#(a+3sqrt5)^2+a-bsqrt5 = 51# if and only if

#a^2+6asqrt5 + 45 +a-bsqrt5 = 51#. Which is equivalent to

#a^2+a+45 +(6a -b)sqrt5 = 51#

Notice that the right hand side is rational, in fact it is an integer. It does not include any #sqrt5#

On the left, since we are told that #a# and #b# are integers, we must have

#a^2+a+45# is an integer and #(6a -b)sqrt5# is irrational unless #6a-b = 0#.

If the sum is to be equal to #51#, then we must have

#a^2+a+45 = 51# #" "# and #" "# #6a -b = 0#

So #a = -3# or #2# #" "# and #" "# #b=6a#