Almost Pythagorean
Number Theory
Level
4
How many ordered triples of positive integer satisfy
Infinitely many
0
Finitely many, from 1 to 100
Finitely many, > 100
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1 solution
Let a = 2 k be any even integer ≥ 4 6 . Then a 2 − 2 0 1 5 is an odd positive number, and we can have it equal to c 2 − b 2 via ( c − b , c + b ) = ( 1 , a 2 − 2 0 1 5 ) .
This gives us c = 2 a 2 − 2 0 1 4 , b = 2 a 2 − 2 0 1 6 .
Thus, we have infinitely many solutions of the form
( a , b , c ) = ( 2 k , 2 k 2 − 1 0 1 3 , 2 k 2 − 1 0 1 2 ) .
Note: Of course, other factorizations of c 2 − b 2 exist, and all of them will work, since the factorization would always be odd * odd.
It seems like everyone is guessing that there are no solutions.