If $\,3\nmid f(n)=2n^2+10n+15\,$ then $\,f(n)\equiv 7\pmod{\!36}$

Question

While attempting to characterize primes of the form $2n^2+10n+15\,$ I discovered

If $\,f(n)=2n^2+10n+15\,$ is prime then $\,f(n)\equiv 7\pmod{\!36}$

I am interested in proving or disproving this .

Attempt: So, I know that we have to have $n\equiv 2\pmod 3$, because, if $n\equiv 1\pmod 3$ then $2n^2+10n+15\equiv 2*1^2+10*1+15\equiv 2+10+15 \equiv 27\equiv 0 \pmod 3$, which is impossible since $p$ can only be divisible by itself and $1$ which would mean that $p=3$, but $p$ is obviously larger than $3$. If $n\equiv 0\pmod 3$, then $2n^2+10n+15\equiv 2*0+10*0+15\equiv15\equiv0\pmod3$, so again we have that a prime number not equal to 3 is divisible by 3, which is impossible, so our only case left is $n\equiv 2\pmod 3$. But I'm stuck here, and I don't think it's that helpful.

Answer 1

Well, as for the $36k+7$ part, if you put $n=3k-1$, then you have: $2(3k-1)^2+10(3k-1)+15=18k^2+18k+7=18k(k+1)+7$. You clearly see that $18k(k+1)$ is divisible by 36 since you have the factor of 18 and $k(k+1)$ is divisible by 2.

Answer 2

In cases like this, it helps to remember that the primes (except the first two) are of the form $p=6k+1$ and $p=6k-1$. So we are looking for solutions of $2n^2+10n+15=6k+1$ and $2n^2+10n+15=6k-1$. So the problem is now reduced to solving a quadratic equation in n with a parameter k. We demand that the discriminant d of these quadratic equations be a square.

The case of $p=6k+1$ has a discriminant $d=12k-3=3*(4k-1)$. For $d$ to be a square, $(4k-1)$ must be of the form $(4k-1)=3*m^2$. For the solutions of the quadratic equation $n_1$ and $n_2$ to be an integer, $m$ must be odd. So the values of $m$ must be $m=1,3,5,7...$, an infinite number of possible values. Then $d=3^2*m^2$ is a square for $m=1,3,5,7,9,11,13...$ and $k=1,7,19,37,61,91,127...$ the corresponding values of $n$ are $n=-1,2,5,8,11,14,17...$, of the form $3j+2$ as shown by Robjohn. Of course there are 2 values of $n$ for each value of $k$ but they lead to the same value of quadratic equation in $n$. The corresponding primes are $p=7,43,223,367,547$ if we skip the values of the pair $k=19,127$ and $n=5,17$ which give the integers $115,763$.

The case of $p=6k-1$ can be handled the same way. I think a simple program (sadly, I cannot code) can convince us that there are many such primes, perhaps even an infinity of them.

For the $36k+7$ part, we start with the expression of the positive solution $n_1=(-5+3m)/2$. We have seen that $m$ must be odd for integer solutions so $m=2j+1$. So $2n_1=-5+3*(2j+1)$ or $n_1=3j-1$. The rest of the steps follows user HeatTheIce's answer. This also justify his setting $n=3j-1$.

For comparison, I will provide the link to a similar question about primes that can be represented as a sum of two consecutive squares and those also could be found by the above method. Those primes are listed in the Online Encyclopedia of Integers sequences.

what prime numbers can be written as the sum of two consecutive squares?

Answer 3

If $n\equiv0\pmod3$, then $2n^2+10n+15\equiv0\pmod3$
If $n\equiv1\pmod3$, then $2n^2+10n+15\equiv0\pmod3$
Therefore, we must have $n\equiv2\pmod3$.

If $n=3k+2$, then $$ \begin{align} 2n^2+10n+15 &=\left(18k^2+54k+36\right)+7\\ &=36\binom{k+2}{2}+7 \end{align} $$

where

Answer 4

Yet another way to get at the same answer is to start out by tackling this problem modulo 6. This is the scenic route.

The squares modulo 6 are 1, 4, 3, 4, 1, 0. Twice that we get 2, 2, 0, 2, 2, 0. Add $10n$ to get 0, 4, 0, 0, 4, 0. And add 15 to get 3, 1, 3, 3, 1, 3. This means that except for $n = 2$ or 5, $2n^2 + 10 + 15$ is an odd multiple of 3 greater than 3 and therefore not prime.

Then, broadening out to modulo 36, we concentrate on these values of $n$: 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35. Squared they are 4, 25, 28, 13, 16, 1, 4, 25, 28, 13, 16, 1, and doubled they are 8, 14, 20, 26, 32, 2, 8, 14, 20, 26, 32, 2.

Add $10n$ to get 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28.

And $28 + 15 = 43 \equiv 7 \bmod 36$.

Answer 5

Below I explain why $\,f(n)=f(-1\!+\!\color{#0af}3\:\!k)\:\!\ \overbrace{\equiv f(-1)}^{\large =\ 7\ \rm in\ OP\!} \pmod{\!2\cdot2\cdot \color{#0af}{3^2}}.\,$ It is essentially a sharpening of a $\rm\color{#0af}{double}$ root test. Here $\color{#0b0}{f'(-1)} = \color{#c00}1\cdot 2\cdot \color{#0af}3$.

Lemma $\ \ \ $ If $\ f(x)\:\! =\:\! a\,x^2 +b\,x + c,\ \ \color{#0b0}{f'(r)} \:\!=\:\! \color{#c00}e\,a\,\color{#0af}q,\:$ $\color{#c00}{{\rm odd}\ \,e},\,$ $\,a,b,c,e,q\in\Bbb Z$
$\quad\ $ then $ \ \ f(r\!+\!k\:\!\color{#0af}q)^{\phantom{|^{^|}}}\!\!\!\equiv f(r)\ \pmod{\!\color{#c00}2\:\!a\:\!\color{#0af}{q^2}}$

$\begin{align}{\bf Proof} \ \ \ f(r\!+\!kq) &= f(r)+\color{#0b0}{f'(r)}\:\!kq + a(kq)^2\ \ \small\text{by Taylor's Theorem}\\[.3em] &= f(r)+aq^2\,\color{#c00}{k(e+k)} \end{align}$

Answer 6

You got $n\equiv2\pmod3$. That means $n\equiv2$ or $n\equiv5$ or $n\equiv8\equiv-1\pmod 9.$

That means $f(n)=2n^2+10n+15\equiv8+20+15=43\equiv7\pmod9$

or $f(n)=2n^2+10n+15\equiv 50+50+15=115\equiv7\pmod9$

or $f(n)=2n^2+10n+15=2(-1)^2+80+15=97\equiv7\pmod 9.$

Furthermore, for all $n,$ $n(n+1)$ is even, so

$f(n)=2n^2+10n+15\equiv2n^2+2n+3\equiv2n(n+1)+3\equiv3\pmod4$.

Therefore, $f(n)\equiv7\pmod{36}$.