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Determinate all the positive integers $x$ such that $x^4+6x^3+11x^2+3x+11$ is a perfect square.

My try

With pure intuition I can say that there is no $ x $ that meets this condition, i tried by brute force some numbers and none worked, but I would like to know if what I'm guessing is right or not, and what would be a way to prove it, because i don't see the way to prove this.

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    $\begingroup$ My rough approach would be: the expression is equal to $(x^2 + 3x + 1)^2 + (-3x + 10)$. So if it's a perfect square, then either $-3x + 10 = 0$ (impossible), or $(x^2 + 3x + 1)^2 + (-3x + 10) \le (x^2 + 3x)^2$, or $(x^2 + 3x + 1)^2 + (-3x + 10) \ge (x^2 + 3x + 2)^2$. But the difference say between $(x^2 + 3x + 1)^2$ and $(x^2 + 3x)^2$ is large compared to $-3x + 10$ for large $x$, so that bounds how large $x$ can be. Leaving only a finite number of $x$ to test manually. $\endgroup$ – Daniel Schepler Feb 14 '18 at 21:01
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    $\begingroup$ @DanielSchepler Or, the same with trying from below: $f(x)=(x^2+3x)^2+2x^2+3x+11$, and the extra $2x^2+3x+11$ is smaller than $(x^2+3x+1)^2-(x^2+3x)=2x^2+6x+1$ except when $x\le 3$ $\endgroup$ – Hagen von Eitzen Feb 14 '18 at 21:08
  • $\begingroup$ Getting a strong deja vu vibe. Fairly sure this is an exact dupe, but may be a near miss. ... Didn't find it right away. G'night. $\endgroup$ – Jyrki Lahtonen Feb 14 '18 at 21:16
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Suppose that exist $n\in\Bbb N$ such that $$x^4+6x^3+11x^2+3x+11=n^2$$

See $\text{mod } 3$ the equation and you can prove that $n^2\not\equiv 2 \text{ mod } 3$ for all $n\in\Bbb N$. In fact, if $n=3k+a$ with $a=0,1,2$ then $n^2=9k^2+6ka+a^2\equiv a^2 \text{ mod } 3$ and this values are $0,1,1$ respectively.

Now,

If $x\equiv 0 \text{ mod } 3$ then $x^4+6x^3+11x^2+3x+11\equiv 0^4+6.0^3+11.0^2+3.0+11\equiv 11\equiv 2 \text{ mod } 3$

If $x\equiv 1 \text{ mod } 3$ then $x^4+6x^3+11x^2+3x+11\equiv 1^4+6.1^3+11.1^2+3.1+11\equiv 1+6+11+3+11\equiv 2 \text{ mod } 3$

If $x\equiv 2 \text{ mod } 3$ then $x^4+6x^3+11x^2+3x+11\equiv 2^4+6.2^3+11.2^2+3.2+11\equiv 16+0+44+0+11\equiv 2 \text{ mod } 3$.

This means that no have solutions

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Hint : $x^{3} + 11x$ is always divisible by 3.

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