Elegance is the key
For how many ordered pairs ( x , y ) of integers x and y is the expression 1 2 x 2 + 2 7 y 2 + 3 6 x y − 1 a perfect square?
The answer is 0.
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2 solutions
Moderator note:
Couldn't your solution just be the last line?
1 2 x 2 + 2 7 y 2 + 3 6 x y − 1 = 3 K + 2 . Hence it is never a perfect square.
After finding a solution, think about if the original approach is the best way to present it.
Nc solution bro
*In Response to the Challenge Master note : *
I have elaborated the solution to make sure that everybody understands it.
1 2 x 2 + 2 7 y 2 + 2 6 x y − 1
= 3 ( 4 x 2 + 9 y 2 + 1 2 x y − 3 1 )
Because the expression is a square, it implies 3 = 4 x 2 + 9 y 2 + 1 2 x y − 3 1 . Then 3 3 1 = 4 x 2 + 9 y 2 + 1 2 x y = ( 2 x ) 2 + ( 3 x ) 2 + 2 ( 2 x ) ( 3 y ) = ( 2 x + 3 y ) 2 → 3 3 1 = 2 x + 3 y Because 3 3 1 is irrational, it cannot be the sum of two integers with integral coefficients, therefore the answer is 0 .
We have
1 2 x 2 + 2 7 y 2 + 3 6 x y − 1
= 3 ( 4 x 2 + 9 y 2 + 1 2 x y ) − 1
= 3 ( 2 x + 3 y ) 2 − 1 = 3 ( ( 2 x + 3 y ) 2 − 1 ) + 2 . --------- Expression (1)
We also have ( 2 x + 3 y ) 2 ≥ 0 .
From Euclids Division Lemma, we know that every perfect square is of the form 3 k or 3 k + 1 , but substituting j = ( 2 x + 3 y ) 2 − 1 in Expression (1),
[ Note : x , y ∈ Z ⇒ j ∈ Z ],
we have that the given expression is of the form 3 j + 2 , and is hence never a perfect square.