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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.