Prime divisors of $k^2+(k+1)^2$

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 $2.$

$$ 6^2 + 7^2 = 85 = 5 \cdot 17 $$

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 $q|x.$