# Let ABCD be a rectangle, and let DM be a segment perpendicular to the plane of ABCD. Suppose that DM has an integer length, and the lengths of MA, MC, and MB are consecutive odd positive integers (in this order). What is the volume of pyramid MABCD?

The volume of the pyramid MABCD is `4sqrt(7).`

Denote the length of DM as `n ` (an integer), the length of DA as `a , ` and DC as `b . ` Then the given conditions become

`sqrt ( n^2 + b^2 ) - sqrt ( n^2 + a^2 ) = 1 , ` `sqrt ( n^2 + a^2 + b^2 ) - sqrt ( n^2 + b^2 ) = 1 .`

The second equation gives `n^2 + b^2 = ( 1 + sqrt ( n^2 + a^2 ) )^2 ` and the sum of the two gives `n^2 + a^2 + b^2 = ( 2 + sqrt ( n^2 + a^2 ) )^2 .`

Subtract the last two equations and obtain `a^2 = 1 * ( 3 + sqrt ( n^2 + a^2 ) ) , ` `a^2 - 3 = sqrt ( n^2 + a^2 ) . ` It is also given that `sqrt ( n^2 + a^2 ) ` is an integer, so `a^2 ` is also an integer.

Further, `a^4 - 6a^2 + 9 = n^2 + a^2 , ` or `a^4 - 7a^2 + 9 - n^2 = 0 , ` a quadratic equation for `a^2 . ` Its discriminant `49 - 36 + 4n^2 = 13 + 4n^2 ` must be a square of an integer. Because 13 is prime, `13 = k^2 - 4n^2 = (k - 2n )(k+2n) ` implies `k - 2n = 1 ; ` in other words, 13 is a difference of two consecutive squares. This is possible only for `n = 3 , ` which implies `a^2 = 7 .`

From `a^2 ` and `n ` we can determine `b^2 = a^2 + 1 + 2sqrt (n^2+a^2) = 8+2sqrt(16)=16, ` and the volume `1/3 abn ` is `4sqrt(7).`

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