# Finding longest sequence such that $4k + 1$ is neither prime nor a perfect square

What is the largest value of $$k$$ such that $$4k + 1$$ produces maximum consecutive terms which are neither primes nor perfect squares and $$k$$ is a natural number.

For example, at $$k = 16$$, we get a sequence of two consecutive terms, 65 and 69 which are neither prime nor square.

One of my friend asked me this question and I am unable to solve it. Can anybody please solve it?

As Tom already explained, you can use the factorials to obtain a sequence of composite non-squares of any length you want.

However, if you only want to go up to a certain bound, your best bet might be to choose the largest $$k$$ less than the bound such that $$k - 1$$ is divisible by two consecutive integers.

For example, 9900 is divisible by 99 and 100. Then $$4 \times 9900 + 1 =$$ $$39601 = 199^2$$. Therefore, with $$k > 9900$$, we can be certain that the next $$k$$ that gives a square for $$4k + 1$$ is $$k = 10100$$, which leads to $$201^2 = 40401$$.

But any of those other $$k$$ between 9900 and 10100 could give a prime, as far as we know right now. When $$k \equiv 1 \pmod 5$$ (it ends in 1 or 6 in base 10) and $$k > 1$$, we can rest assured $$4k + 1$$ is a multiple of 5, as is, for example, 39605. And we get lucky with 36909 being a multiple of 3.

I give Mathematica the command Select[Range[9890, 10100], PrimeQ[4# + 1] &] and it answers that $$k = 9927$$ plugged into $$4k + 1$$ gives a prime, as well as a bunch of other numbers (this works in Wolfram Alpha, too).

Applying Differences[%] (you'll have to be a bit more verbose in Wolfram Alpha, but not necessarily with too much typing) we see the largest difference is 32, but that includes the square 39601 produced by $$k = 9900$$.

So the longest sequence up to your stated bound of 10000 might indeed be of length 26, but I don't feel like checking $$k < 9000$$. Not that it would be too difficult in Mathematica or even Wolfram Alpha, it just takes a bit more mental effort than I'm willing to expend on this problem.

Given any $$n>1$$, if $$M=\left((4n+1)!\right)^2+5$$ is of the form $$4k+1.$$

And $$M,M+4,M+8,\dots,M+4(n-1)$$ are not prime, and they are not squares, because the next largest square is $$\left((4n+1)!+1\right)^2,$$ which is $$M+2(4n+1)!-4$$ and it is easy to see that $$2(4n+1)!-4>4(n-1).$$

So there is no longest sequence.

You can replace $$(4n+1)!$$ with the product $$2\prod_{j=1}^{n}(4j+1)$$ to get a smaller $$M,$$ but still it is by far likely not to be the smallest $$M$$ such that $$M,\dots,M+4(n-1)$$ have this property.

• What if set the condition k<10000?? – The Demonix _ Hermit Sep 22 '19 at 16:04
• With an upper bound, it is just painful calculation. @TheDemonix_Hermit – Thomas Andrews Sep 22 '19 at 16:09
• My computer calculated and found it to be 26.Can It be true?? – The Demonix _ Hermit Sep 22 '19 at 16:14
• $k = 26$? I doubt it. I'm writing up an answer... – Robert Soupe Sep 22 '19 at 21:00
• @Mike you have that $M+4j$ is divisible by $4j+5$ for $j=0,1,\dots,n-1.$ – Thomas Andrews Sep 23 '19 at 19:39