Given
a^2+15226=c^2,
How many solutions (a,c) are there such that both numbers are positive integers?
Please comment upon the unexpected result.
<o:p> </o:p>
since 15226= 2*7613 and 7613 is a prime number - let's solve for a general case:
(c+a) *(c-a) =2*p , p an odd prime
Solving for a & c there are 2 possible integer factorizations :
c+a =2p c-a=1 or
c+a=p , c-a=2
In both sets of equation there are no integer solutions:
c+a and c-a must be of the same parity, i.e. both odd or both even.
see my "spoiler" post
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Edited on March 1, 2018, 4:15 am