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

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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

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