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

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?

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## Comments

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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

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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

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

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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?

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This is not correct @Krishna Ar

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why

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Consider

\(n^6+n^5\)

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

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