Prove that

i) the product of 4 consequtive positive integers plus 1 must be a perfect square

and therefore show that

ii) for any positive integer n, n^4+2n^3+2n^2+2n+1 is not a perfect square.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

i) The product of 4 consecutive integers + 1 can be written as

`n(n+1)(n+2)(n+3) +1`

Multiplying this out we get


`(n^2+n)(n^2 + 5n+6) + 1 `

`= n^4 + ([5n^3],[n^3]) + ([6n^2],[5n^2]) + 6n + 1`

` ` `= n^4 + 6n^3 + 11n^2 + 6n+1`.

If this can be factorised into two identical factors then the factors will be of the form

`n^2 + an + b`

Since the unit term in the expression is 1 we have that `b=1` . Multiplying out the factors we get

`(n^2+an+1)(n^2+an+1) `

`= n^4 + ([an^3],[an^3]) + ([n^2],[a^2n^2],[n^2]) + ([an],[an]) +1 `  `= n^4 + 2an^3 + (a^2+2)n^2 + 2an +1`

Matching the coefficient in `n^3` to that in the expression of interest we have that `a` must equal 3. Checking the other coefficients, they match up: `(a^2+2) = 11` , `2a = 6`. Therefore, the expression is a perfect square and can be written as `(n^2+3n+1)^2` .

ii) This second expression `n^4 + 2n^3 + 2n^2 + 2n+1` can only be a perfect square if it is consistent with the equation above, namely for some `a` it equals

`n^4 + 2an^3 + (a^2+2)n^2 + 2an +1`   (since again `b=1` here)

Matching up the coefficients in `n^3` we require then that `a=1`. However with this value for `a` the coefficients in `n^2` don't match: `(a^2 + 2) = 3 != 2`.

Therefore this second expression cannot be a perfect square for positive integers `n`.

Answer: i) the expression is a perfect square ii) the expression isn't a perfect square

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial