Prove it for perfect square

Prove that for no integer $n, n^6 + 3n^5 - 15n^3 + 4n^2 + 12n + 3$ is a perfect square.

#NumberTheory #PerfectSquare

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

Sorry for late response,

Its very easy to prove that for n=even number , let n=2k , k - integer,

$n^6 + 3n^5 - 15n^3 + 4n^2 + 12n + 3$

$= 64k^6 + 96k^5 - 80k^4 - 120k^3 + 16k^2 + 24k + 3$

$= 4(something) + 3$ , it can't be perfect square for any even integer

For odd , n = 2k +1 ,

$n^6 + 3n^5 - 15n^3 + 4n^2 + 12n + 3$

$(2k+1)^6+ 3(2k+1)^5 .... + 3$

We can see that $(2k+1)^6 \equiv (12k + 1) \pmod{4} \equiv 1 \pmod{4}$

$3(2k+1)^5 \equiv (30k + 3) \pmod{4} \equiv (2k + 3) \pmod{4}$

$5(2k+1)^4 \equiv (40k + 5) \pmod{4} \equiv 1\pmod{4}$

$15(2k + 1)^3 \equiv (90k + 15) \pmod{4} \equiv (2k +3) \pmod{4}$

$4(2k+1)^2 \equiv 0 \pmod{4}$

$12(2k+1) \equiv 0 \pmod{4}$

or $n^6 + 3n^5 - 15n^3 + 4n^2 + 12n + 3 \equiv (1 + 2k + 3 + 1 + 2k +3 +3) \pmod{4} \equiv (4(k+2) + 3) \pmod{4}$

thus ,the expression $= 4(something) + 3$

Thus it can't be perfect square for any odd integer as well

U Z - 6 years, 5 months ago

I was thinking can this also be a solution.

Let $f(x)=n^6 + 3n^5 - 15n^3 + 4n^2 + 12n + 3$ . Then $f(1)=3, f(2)=3, f(3)=723, f(4)=5043$ . In a similar way we see that all of the values end with 3 as the units digit while in a perfect square the units digit ends with $0,1,4,5,6$ or $9$ . so for no integer n $f(x)$ is a perfect square

Akhilesh Prasad - 6 years, 6 months ago

Actually this was a question that I got in the KVS -Junior Mathematics olympiad. Pretty much did the same way you explained it.

Akhilesh Prasad - 6 years, 6 months ago

A perfect square with degree 6, ought to be a perfect-sixth. Since , we can clearly see using either Binomial completion of the sixths or by Simple Congruency case-work, that it can be a sixth, we can conclude. ~~~. Whatsay?

Krishna Ar - 6 years, 8 months ago

This is not correct @Krishna Ar

Nathan Ramesh - 6 years, 6 months ago

why

Krishna Ar - 6 years, 6 months ago

@Krishna Ar – It does not necessarily have to be a square for all integers n for the problem to be satisfied

Consider

$n^6+n^5$

It is not a perfect 6th but for n=0 it is a perfect square

Nathan Ramesh - 6 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$