After creating a list of prime numbers under 10000, the first 28 powers of 2, and the first 15000 perfect squares, I ran the following python code, and came up with
one solution.
P=11 ... (
X, Y) = (
4, 7) and (
8, 17)
2^4 + 3*11 = 16 + 33 = 49 = 7^2
2^8 + 3*11 = 256 + 33 = 289 = 17^2
for p in primes:
total=0
values=[]
for x2 in powers_of_two:
y2=x2+3*p
if squares.count(y2):
total+=1
values.append([powers_of_two.index(x2)+1,squares.index(y2)+1])
if total==2:
print(p,values)
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Posted by Justin
on 2010-01-24 16:12:14 |