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which equation supports the given conjecture the sum of the squares of two consective integers is an odd integer

Sagot :


As soon as I saw the word "Which...", I knew that there's a list of choices
to go along with this question, and I was pretty sure that when I opened
the question, the list wouldn't be there.  Thanks for not disappointing me.

One integer can be  'x' .  Then the next consecutive integer is  (x+1).

Let's look and see what the sum of their squares looks like:

(x)² = x²

(x+1)² = x² + 2x + 1

Their sum:  (x²)  +  (x² + 2x + 1) = 2x² + 2x + 1 .

That's the expression we need.  Look at it closely.
No matter what  'x' is ...

   --  (2x²)  is definitely an even number
and
   --  (2x)  is definitely an even number
so
   --  (2x² + 2x)  is definitely an even number
and
   --  (2x² + 2x + 1)  (the sum of the integers) is definitely an odd number.

QED


Answer:

bruh add the choices

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