P is a positive integer such that each of 2P + 1 and 3P + 1 is a perfect square.

Prove that P is divisible by 40

2P+1 is odd, so the square of an odd number
2p+1=(2k+1)² => p=2(k²+k) => p is even
So, 3P+1 is also odd perfect square
let 2P+1 = a² and 3P+1 = b²
i)P=(3P+1)-(2P+1)=(b-a)(b+a)
One of b-a, b+a is divisible by 4 and the other by 2.
So, P is divisible by 8.

ii)5P+2=a²+b² --- (1)


for any perfect square x, x mod 5 =(0,+1,-1)mod 5
For Your Info on modulo addition:
a²+b²=0 mod 5 when (a²,b²)=(0,0),(+1,-1),(-1,+1) mod5

a²+b²=1 mod 5 when (a²,b²)=(+1,0),(0,+1) mod 5
a²+b²=2 mod 5 when (a²,b²)=(+1,+1) mod 5
a²+b²=3 mod 5 when (a²,b²)=(-1,-1) mod 5
a²+b²=4 mod 5 when (a²,b²)=(-1,0),(0,-1) mod 5


For equation (1) take modulo 5 on both sides
a²+b² mod 5 = 2mod5
so, a²=1 mod 5 and b²=1 mod 5
So, 2P+1 mod 5=1 => 2P mod 5=0 => P is divisible by 5.

P is divisible by 5 and 8. So, P is divisible by 40.

Posted by Praneeth on 2008-05-13 23:16:31