2^48- 1

=(2^24+ 1)(2^ 12+ 1)(2^6+ 1)(2^3+ 1)(2^3- 1)

=(2^24+ 1)(2^ 12+ 1)*65*9*7

=(2^24+ 1)(2^ 12+ 1)*65*63

So, these two integers are 63 and 65.

Binomial numbers a, b and c satisfy A+ 18 = B+ 14 = C+35.

Find the value of a 2+b 2+c 2-ab-BC-ca.

a+ 18=b+ 14=c+35

a-b=-4，b-c=2 1，a-c= 17

a^2 + b^2 + c^2 - ab - bc - ca

= 1/2(2a^2+2b^2+2 c^2-2ab-2bc-2ca)

= 1/2(a-b)^2+ 1/2(a-c)^2+ 1/2(b-c)^2

= 1/2( 16+44 1+289)

=373

Three experiments have proved that the sum of the products of four consecutive natural numbers 1 must be a complete square number.

Prove:

Let these four connected natural numbers be (n-2), (n- 1), n and (n+ 1) respectively.

(n-2)*(n- 1)* n *(n+ 1)+ 1

=(n^2-n)(n^2-n-2)+ 1

=(n^2-n)^2-2(n^2-n)+ 1

=(n^2-2n- 1)^2

N is a natural number, so n 2-2n- 1 is a natural number.

Therefore, the sum of the products of four continuous natural numbers and 1 must be a complete square number.