Let each of
x_{1}, x_{2}, x_{3}, …, x_{777}, y_{1}, y_{2}, y_{3}, …, y_{777} be an arbitrary nonzero integer number.
Consider the product
P = (2x_{1}^{2} +3y_{1}^{2}) *
(2x_{2}^{2} +3y_{2}^{2}) * (2x_{3}^{2} +3y_{3}^{2}) * ...* (2x_{777}^{2} +3y_{777}^{2}).
Prove: P cannot be a square number.
(In reply to
re(3): Poossible solution...very creative! by Ady TZIDON)
I thought someone else might puzzle this out, but apparently not.
The crucial step is to assume a.b.c.d. such that a^2+6b^2 =2c^2+3d^2;
Then 6b^23d^2 =2c^2a^2 and so 3(2b^2d^2) =2c^2a^2;
but it is clear that there are no such numbers; (a^22b^2)(c^22d^2) = (2bd+ac)^2  2(ad+bc)^2; the product of two numbers of form (x^22y^2) is another number of the same form.
But it is easily computed that 3 is not of that form (though 9 is).
So the two sequences of numbers are distinct. Hence if we find the squares in a^2+6b^2, as already explained, we can be sure that there are no squares in 2c^2+3d^2.
Edited on March 4, 2017, 8:42 am

Posted by broll
on 20170304 08:40:33 