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 Never a square (Posted on 2017-02-26)
Let each of x1, x2, x3, …, x777, y1, y2, y3, …, y777 be an arbitrary non-zero integer number.
Consider the product

P = (2x12 +3y12) * (2x22 +3y22) * (2x32 +3y32) * ...* (2x7772 +3y7772).

Prove: P cannot be a square number.

 No Solution Yet Submitted by Ady TZIDON No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 re(4): Poossible solution...very creative! Comment 5 of 5 |
(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^2-3d^2 =2c^2-a^2 and so 3(2b^2-d^2) =2c^2-a^2;

but it is clear that there are no such numbers; (a^2-2b^2)(c^2-2d^2) = (2bd+ac)^2 - 2(ad+bc)^2; the product of two numbers of form (x^2-2y^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 2017-03-04 08:40:33

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