Prime divisors of $k^2+(k+1)^2$ It seems that the prime divisors of : $k^2+(k+1)^2$ are of the form $4k'+1$, and that they can be only one prime divisor of $k^2+(k+1)^2$ of the form : $4k''+1$.
Yet I don't know how to prove this result, and how to study in general the prime divisors of : $k^2+(k+1)^2$. 
 A: For any prime  $q \equiv 3 \pmod 4,$ we have Legendre symbol $(-1|q) = -1.$ That is to say, there is no integer $t$ such that $$  t^2 \equiv -1 \pmod q
$$

The first supplement[6] to the law of quadratic reciprocity is that if
  p ≡ 1 (mod 4) then −1 is a quadratic residue modulo p, and if p ≡ 3
  (mod 4) then −1 is a nonresidue modulo p.

https://en.wikipedia.org/wiki/Quadratic_residue#Prime_modulus
you asked something about this earlier but then deleted it. If there is a prime $q \equiv 3 \pmod 4,$ and $q | (u^2 + v^2),$ then both $q | u$  and $q | v;$ in this situation, it follows that $q | \gcd(u,v)$ and $\gcd(u,v) \neq 1.$
Well, $\gcd(k, k+1) = 1.$ Therefore $k^2 + (k+1)^2$ cannot be divisible by any prime  $q \equiv 3 \pmod 4.$
Your number is also odd, so not divisible by $2.$
$$  6^2 + 7^2 = 85 = 5 \cdot 17 $$
A: I know that I have proved this on this site, cannot find it. This time I have saved the link to my text file on my home computer, topics.txt
this is  Prime divisors of $k^2+(k+1)^2$
Let 
$$ f(x,y) = a x^2 + b xy + c y^2   $$ be a quadratic form with integer coefficients $a,b,c.$ Define the "discriminant"
$$ \Delta = b^2 - 4 a c.  $$
We demand that $\Delta \neq 0,1,4,9, \ldots$ That is, if $\Delta $ is non-negative, we require that it not be a square. 
Proposition: let $q$ be an odd prime with Legendre symbol
$$  (\Delta | q) = -1. $$
That is, $q$ does not divide $\Delta,$ and there is no integer $t$ such that $$ t^2 \equiv \Delta \pmod q. $$  IF we have integers $x,y$ such that
$$  a x^2 + b xy + c y^2 \equiv 0 \pmod q, $$
THEN
$$ q|x \; \; \; \mbox{AND} \; \; \; q | y. $$ Therefore we actually have $  a x^2 + b xy + c y^2 \equiv 0 \pmod {q^2} $
PROOF: The integers $\pmod q$ make a field. As we required $q$ odd, we know $4$ is invertible in $\mathbb Z/ q \mathbb Z.$ Next, we cannot have $q |a,$ because then we would have $(\Delta|q) = (b^2|q);$ we would have $b \neq 0 \pmod q,$ so we would get $(\Delta|q) = (b^2|q)= 1.$ But this would contradict the hypothesis $(\Delta|q) = -1.$
We therefore have  $4a \neq 0 \pmod q$ and the equivalent hypothesis
$$  4a^2 x^2 + 4ab xy + 4ac y^2 \equiv 0 \pmod q.  $$
$$  4a^2 x^2 + 4ab xy + b^2 y^2 - b^2 y^2 + 4ac y^2 \equiv 0 \pmod q.  $$
$$ \left( 4a^2 x^2 + 4ab xy + b^2 y^2 \right) - \left(b^2 - 4ac \right) y^2 \equiv 0 \pmod q.  $$
$$ \left( 4a^2 x^2 + 4ab xy + b^2 y^2 \right) - \Delta y^2 \equiv 0 \pmod q.  $$
$$ \left( 2ax + by \right)^2 - \Delta y^2 \equiv 0 \pmod q.  $$
$$ \left( 2ax + by \right)^2 \equiv \Delta y^2  \pmod q.  $$
We finally get to the prof by contradiction.
ASSUME $y \neq 0 \pmod q.$ In that case, $y$ has a multiplicative inverse in the field, call it $w,$ so $yw \equiv 1 \pmod q.$ Then $y^2 w^2 \equiv 1 \pmod q,$
$$ \left( 2ax + by \right)^2 w^2 \equiv \Delta y^2  w^2 \pmod q.  $$
$$ \left( (2ax + by \right) w)^2 \equiv \Delta  \pmod q.  $$
This is a CONTRADICTION of $(\Delta|q) = -1.$ We have shown that $y \equiv 0 \pmod q.$ In the original
$$  a x^2 + b xy + c y^2 \equiv 0 \pmod q, $$ this gives
$$  a x^2 \equiv 0 \pmod q. $$
Since we showed that $a \neq 0 \pmod q,$
$$   x^2 \equiv 0 \pmod q, $$ finally
$$   x \equiv 0 \pmod q. $$ Together, we have both $q|y$ and $q|x.$
A: $k^2+(k+1)^2=2k^2+2k+1=\frac12((2k+1)^2+1)$.
If a prime $p$ divides $2k^2+2k+1$ it is odd, and $(2k+1)^2\equiv-1\pmod p$
and so $p\equiv1\pmod 4$.
Conversely, if the prime  $p\equiv1\pmod 4$ then there is $a$ with $a^2\equiv-1\pmod p$. We may assume $a=2k+1$ is odd (otherwise consider $a+p$) and then $p\mid(2k^2+2k+1)$.
