2019 and sum of squares
How many triples ( x , y , z ) satisfy the equation x 2 + y 2 = z 2 + 2 0 1 9 where x , y , z are integers
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1 solution
To make the argument more simple and clear, we can set x − z = 1 and just argue for x , y .
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Yes, that would also work. Thanks!
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How would that work?
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@Mr. India – If x − z = 1 , then x + z is some odd integer, and ( x + z ) ( x − z ) = x + z . x + z can also be ANY odd integer. Then we see that if 2 0 1 9 − y 2 can equal an odd integer if y is odd. Therefore, for every odd y , there is some x and z such that x + z = 2 0 1 9 − y 2
x 2 + y 2 = z 2 + 2 0 1 9 x 2 − z 2 = 2 0 1 9 − y 2 ( x + z ) ( x − z ) = 2 0 1 9 − y 2
Here, we can see that two numbers multiplied together to get some number 2 0 1 9 − y 2 . There are an infinite possibilities of integers that 2 0 1 9 − y 2 can make, and an infinite many of those are not prime, and so there are an infinite number of solutions for this problem. (Note that there are a few other requirements, such as that x + z and x − z are either both positive or negative. But in the long run, this does not change the number of solutions to finite.)