Math Is Fun Forum

I've done the first part using induction. The polynomials I chose were:

However, the last part is a bit tricky.

I can't see how to write this in the form

Can anyone help? (Small hints preferred) Thanks.

The four digit integer aabb is a perfect square. Find a and b.

n^2 = 1000a + 100a + 10b + b = 1100a + 11b = 11(99a + a + b)

n^2 is divisible by 11, so it must also be divisible by 11^2. Therefore 11|(a+b), but since 1 < a + b < 18, a + b = 11.

I got this far, but wasn't really sure how to continue. My book's solution then says:

"Now since n^2 is a square, b cannot be 2,3,7 or 8. Also, b cannot be 5, since the square would end in 25, not 55"

I don't get this :S

Thanks.

This is annoying me.

Prove that for n>3, 1! + 2! + 3! + ... + n! is never square.

I've managed to prove it by considering modulo 5. All squares are either 1 or 4 (mod 5). n! = 0 (mod 5) for n≥5. Therefore the sum = 1! + 2! + 3! + 4! = 3 (mod 5), and so it can't be square.

However, I first attempted it by considering modulo 4. All squares are either 0 or 1 (mod 4). But the sum is odd, and so the square would have to be 1 (mod 4). n! = 0 (mod 4) for n≥4. Therefore the sum = 1! + 2! + 3! = 1 (mod 4). It could therefore be square??

Thanks.

Prove that if