Prove that primes of the form $2n^2+10n+15$ are also of the form $36k+7$

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 divisable with 36 since you have the factor of 18 and $k(k+1)$ is divisable with 2.

By the Bunyakovsky conjecture we expect that there are infinitely many primes of the form $2n^2+10n+15$. However, this is still an open conjecture; the case $n^2+1$ is probably the most popular one right now.

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?