If A=m3×n×p2, where m, n and p are prime numbers, which of the follo

1.	If A=m3×n×p2, where m, n and p are prime numbers, which of the following is a perfect square?
Answer» Answer: More generally, we show that the only integers n for which 1+n+n2+n3+n4 is a square are −1 , 0 , and 3 . Let F(n)=4n4+4n3+4n2+4n+4 , n∈Z . From f(n)=(2n2+n)2+3(n+23)2+83 , we have f(n)>(2n2+n)2 . From f(n)=(2n2+n+1)2−(n+1)(n−3)…(1) , we have f(n)<(2n2+n+1)2 EXCEPT when −1≤n≤3 . Since f(n) lies between (2n2+n)2 and (2n2+n+1)2 except when n∈{−1,0,1,2,3} , f(n) is a square implies n∈{−1,0,1,2,3} . From eqn. (1) , f(−1) and f(3) are squares, as is f(0)=4 , whereas f(1)=42+4=20 and f(2)=112+3=124 are not. Since f(n) is even, f(n) is a square if and only if 14f(n) is a square. THEREFORE we have shown that for n∈Z 1+n+n2+n3+n4 is a square ⇔n∈{−1,0,3} . ■

If A=m3×n×p2, where m, n and p are prime numbers, which of the following is a perfect square?

Answer»

Answer:

More generally, we show that the only integers n for which 1+n+n2+n3+n4 is a square are −1 , 0 , and 3 .

Let F(n)=4n4+4n3+4n2+4n+4 , n∈Z . From

f(n)=(2n2+n)2+3(n+23)2+83 ,

we have f(n)>(2n2+n)2 .

From

f(n)=(2n2+n+1)2−(n+1)(n−3)…(1) ,

we have f(n)<(2n2+n+1)2 EXCEPT when −1≤n≤3 .

Since f(n) lies between (2n2+n)2 and (2n2+n+1)2 except when n∈{−1,0,1,2,3} , f(n) is a square implies n∈{−1,0,1,2,3} .

From eqn. (1) , f(−1) and f(3) are squares, as is f(0)=4 , whereas f(1)=42+4=20 and f(2)=112+3=124 are not.

Since f(n) is even, f(n) is a square if and only if 14f(n) is a square. THEREFORE we have shown that for n∈Z

1+n+n2+n3+n4 is a square ⇔n∈{−1,0,3} . ■