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Proof Verification: there are infinite values of $a$ such that $n^4 + a$ is composite for all $n \in Z$

Let $n^4 + a = (n^2+xn+p)(x^2+yn+q)$. Knowing that the cubic and quadratic coefficients must be zero, we can show that $x = -y, p+q=x^2,$ and $pq = a$.

One immediate solution is $p = 1, q = 3, (x, y) = \pm2,$ for which $a = 3.$

Assume that there exists a finite, but more than zero, number of $a's$ generated in this manner, and the highest $a$ is generated from values $p, q, x,$ and $y$.

Then $x = -y, p+q=x^2,$ and $pq = a_{highest}$

Let $p_1 = q_1 = 2x^2$. Then $p_1 + q_1 = (2x)^2$. Let $x_1 = 2x$ and $y_1 = -2x$.

Since $p_1 > p$ and $ q_1 > q$, $a_{new} = p_1q_1 > a_{highest}$, which contradicts our assumption. Thus there must be an infinite number of $a$ satisfying the conditions.

Is my proof correct?

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Your idea of identifying the coefficients is right, the details are not. After you have identified the coefficients, it should be straightforward. One has to check that neither quadratic can be equal to $\pm 1$. –  André Nicolas Jan 14 '14 at 1:49

5 Answers 5

For any integer $k>1$ we have $$n^4+4k^4=(n^2+2kn+2k^2)(n^2-2kn+2k^2)=((n+k)^2+k^2)((n-k)^2+k^2),$$ which is composite for all $n$, being the product of two integers greater than one.

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By what method do you propose to prove that it is always composite? –  Bill Dubuque Jan 14 '14 at 3:15
@BillDubuque Isn't it evident? –  Pedro Tamaroff Jan 14 '14 at 3:18
Well, it fails for $k=1$ since one of the factors can become 1. –  Byron Schmuland Jan 14 '14 at 3:20
Ah, an actual correct solution! –  Igor Rivin Jan 14 '14 at 3:22
Thanks for editing it to make it clearer. I worried that step would not be clear to some readers. Now I can stop editing my answer since it is basically the same. –  Bill Dubuque Jan 14 '14 at 3:26

Your proof seems to confuse composite numbers and reducible polynomials (and in fact, $n^4+3$ is irreducible, so there is something wrong with your computation -- why is the leading coefficient of the second factor $x^2?$). Also, for $n=2,$ the number $2^4+3 = 19$ is not composite.

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Mr. Igor Rivin has point out the error in you proof. We may approach for it by Fermat's Little Theorem.

We have $n^4 \equiv 1$ (mod $5$). Which gives $5| n^4 -1$, i.e. $n^4 -1$ is composite. Take $a \equiv 1$ (mod $5$). You shall get countably infinite $a$ s.t. $n^4 + a$ is a composite number.

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Umm, not so fast, what if $n$ is divisible by $5?$ –  Igor Rivin Jan 14 '14 at 2:00
@IgorRivin If $n\neq 0\mod 5$, then $n^4=1\mod 5$. That's what F$\ell$T says. –  Pedro Tamaroff Jan 14 '14 at 3:10
@PedroTamaroff Yes, but $n^4+a$ has to be composite for ALL $n.$ –  Igor Rivin Jan 14 '14 at 3:20
If $n$ is divisible by $5$ we shall choose $a$ as multiple of 5. Then $n^4 + a$ is composite. –  Dutta Jan 14 '14 at 5:21

I am sorry I had to rethink the solution.

By the Division algorithm we can prove that the square of any integer is of the form $3k$ or $3k +1$. An extension would give that $n^4 = 3k $ or $3k+1$ for some integer $k$.

If $n^4 = 3k$ then for any $a$ in $\{ 3p \ |$ $p$ is an integer $\}$ the expression $n^4 +a$ is composite given that $p$ is non-zero.

Similarly $n^4 = 3k + 1$ then any $a$ in $\{ 3p +2 \ |$ $p$ is an integer $\}$ will render the expression $n^4 +a$ composite.

Both sets mentioned above have countably infinite many elements.

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Yes, but their intersection is empty, so this does not answer the question. –  Igor Rivin Jan 14 '14 at 3:21
Yes but both cases are also disjoint isn't it? Is it necessary we give a general set for both cases. Depending on the value of $n$ we choose the set. One of which will solve our problem and will have infinitely many elements. Although Byron Schmuland's solution above is much more elegant –  Ishfaaq Jan 14 '14 at 3:25
YES, the question asks for an $a$ which works for ALL $n.$ Neither of yours do. –  Igor Rivin Jan 14 '14 at 3:30

See here: This is obvious that n^4+4k^4=(n^2+2kn+2k^2)(n^2−2kn+2k^2)=((n+k)^2+k^2)((n−k)^2+k^2) the above written expression is always composite for all n and k>1. As you can see the expression : n^4 + 4k^4 can be factorised in this manner....so for any values of k>1 the number is composite.

So, the form of the number 'a' should be of form : 4k^4 And easily it is observed that there are infinitely many number which is of form : 4(k^4).

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