2
$\begingroup$

So, I tried solving that by $$n^4-4n^3+14n^2-20n+10=x^2\\10=x^2-a^2, a^2=n^4-4n^3+14n^2-20n+10\\10=(x+a)(x-a)$$ but I couldn't find any integers when I solved it

$\endgroup$
4
  • 3
    $\begingroup$ Let $m = n-1$ and write things in terms of $m$. That should make things easier to see. $\endgroup$ May 24, 2020 at 19:00
  • 1
    $\begingroup$ What about $n=1$ and $n=3$? $\endgroup$ May 24, 2020 at 19:00
  • 2
    $\begingroup$ Hint: $x^2=(n^2-2n+5)^2-15.$ $\endgroup$
    – user436658
    May 24, 2020 at 19:15
  • 1
    $\begingroup$ I think (1,1),(3,7) are the only solutions, up to sign $\endgroup$
    – user581023
    May 24, 2020 at 20:15

3 Answers 3

4
$\begingroup$

It always helps to form squares from the biggest power and is a good strategy: $$n^4-4n^3+14n^2-20n+10=n^2(n^2-4n+4)+10n^2-20n+10=\\ n^2(n-2)^2+10(n-1)^2=(n(n-2))^2+10(n-1)^2=((n-1)^2-1)^2+10(n-1)^2=\\ (n-1)^4-2(n-1)^2+1+10(n-1)^2=(n-1)^4+8(n-1)^2+1=\\ ((n-1)^2+4)^2-15=x^2$$ I think this quite large hint will make it a bit easier to solve it.

Just as a note: it just happens we can make a nice square, that is not always the case!

$\endgroup$
1
$\begingroup$

Let f(n) = n^4 - 4*n^3 + 14*n^2 - 20*n + 10

As Daniel Fischer mentioned you can apply the substitution $m = n-1$ to get f(m+1) = m^4 + 8*m^2 + 1

complete the square, we want to solve $$(m^2 + 4)^2 - 15 = a^2.$$ subtract one from both sides an use the difference of squares factorization $$m^2 (m^2 + 8) = (a - 1)(a + 1)$$

lets put $b = a-1$ and $h = n^2$, we can work on this factored form of the problem: $$h (h + 8) = b (b+2)$$


Clearly there are 4 solutions attained by making both sides zero $(h,b) = (0,0), (0,-2), (-8,0), (-8,-2)$.

Let's restrict to positive integer solutions from now on since $h$ is a square it must be positive. Since $8 > 2$ A solution must have $h < b$. Put $L(h) = h (h + 8)$ and $R(b) = b (b+2)$.

  • $L(h) < B(h)$
  • $L(h) > B(h+6)$

therefore any solution will have $b = h+1$ or $h+2$ or $h+3$ or $h+4$ or $h+5$. With this in mind let's substitute in $b = h+k$:

So we only need to find small $k$ integer solutions of the following linear equation now

$$0 = (-2k + 6)h + (-k^2 - 2k + 1)$$

checking each case we quickly find there are no solutions at all.


In conclusion the only solutions are the four listed from making both sides zero.

$\endgroup$
1
  • 1
    $\begingroup$ With $b=m^2+4,$ we have $b^2=15+a^2+15>a^2$ so $(b-1)^2\ge a^2$.... So $b\le 8$ because $b\ge 9\implies b^2=(b-1)^2+2(b-1)+1\ge a^2+2(b-1)+1\ge a^2+17$ .... So $m^2+4\le 8.$ $\endgroup$ May 25, 2020 at 6:42
0
$\begingroup$

Let us expand $(x^2+ax+b)^2 = (x^2+ax+b)(x^2+ax+b)$:

$$x^4 + (2a)x^3 + (a^2+2b)x^2 +(2ab)x+b^2$$

We want this to be as close as possible to the given expression, so we need to match the coefficients to the terms in the expansion. Starting from the coefficient after $x^4$ which is $x^3$, when $2a = -4 \Rightarrow a = -2$, we obtain $x^4 - 4x^3 + (4+2b)x^2 - (4b)x + b^2$.

Similarly, when $4+2b=14 \Rightarrow b = 5$, then we have $x^4 - 4x^3 + 14x^2 - 20x + 25$, which is exactly $15$ more than the given expression.

Therefore $x^2 = (n^2 - 2n + 5)^2 - 15$, and you can continue by using the difference of squares method explained in rain1's answer.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .