A problem from the Russian Math Olympiad

 

Prove that the product of four consecutive integers plus one is the square of an integer.

 

Proof.

 

1. Here is an algebraic equality that will prove the assertion.

 

(a - b)*a*(a + b)*(a + 2*b) + b4 = (a2 + a*b - b2)2

 

In order to prove this equality, check that each side is equal to

 

a4 + 2*a3*b - a2*b2 - 2*a*b3 + b4

 

Remark.

The use of this expression instead of a*(a+b)*(a+2*b)*(a+3*b) + b4 leads to slightly simpler algebraic computations.

 

2. The required proof follows from the above equality for b = 1, and a

being any given integer:

 

(a-1)*a*(a+1)*(a+2) +1 = (a2 + a*1 - 12)2

 

You can check this with different values for a > 2!

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