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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.

2 years, 1 month ago

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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 · 1 year, 9 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 · 1 year, 11 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. · 1 year, 11 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? · 2 years ago

This is not correct @Krishna Ar · 1 year, 10 months ago

why · 1 year, 10 months ago

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 · 1 year, 10 months ago