‘PRIMES AND FACTORS’ 59

In order that an integral value for » may be derived from the
above, it is essential that a2 —2a+4./N+1 be a perfect square,
with a square root of the form (a+x), whence

a?—2a+4N+1=(a+x)%2=a%+2ax+x>
or 2ax +2a=(4N+1) —x?
Sorting out terms and changing signs where necessary, so

o o

2(x+1)
Therefore, if there is any integral solution of
n?+(a—1)n—N=o
(4N 1) —x?

then 2¢=- i and from this it will be seen that 2 is
x+1
a factor of (4N +1)—x2 so that this latter number must be
even. Now (4N -+1) is odd and therefore x2 must also be
odd. From the same evidence x-+1 is even and therefore
(4N +1) —x2 must be a multiple of 4. Furthermore, in order
to satisfy the solution required, the integral number, if any,
N+41) —x2 i ;
represented by _<47<j°—+)) must be even as it 1s twice as
xX+1

great as the hypothetical a, which is also required to be
integral.

But there can only be a real value for a and = if ¥ can be
represented as a sum of a series; that is if it is composite.
Therefore if an odd value can be found for x (other than

N4 oah i !

(4 % +)1 —— is an even integer, then N is com-
posite. Otherwise, /N is prime.

Supposing the number to be composite, it is then an easy
matter to evaluate 7, which is itself one of the factors. In order
to arrive at the above identity, we substituted (a+x) for

vV (a—1)244N, so that reverting to the original equation for
4N, g g q

x=1) so that

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