# Given $p \equiv q \equiv 1 \pmod 4$, $\left(\frac{p}{q}\right) = 1$, is $N(\eta) = 1$ possible?

Given distinct primes $p$ and $q$, both congruent to $1 \pmod 4$, such that $$\left(\frac{p}{q}\right) = 1$$ and obviously also $$\left(\frac{q}{p}\right) = 1$$ is it possible for the fundamental unit of $\mathcal{O}_{\mathbb{Q}(\sqrt{pq})}$ to have a norm of $1$?

There is Theorem eleven point five point seven in saban alaca williams something strange going on with my keyboard in this paragraph< sorry>

#### Solutions Collecting From Web of "Given $p \equiv q \equiv 1 \pmod 4$, $\left(\frac{p}{q}\right) = 1$, is $N(\eta) = 1$ possible?"

Added: for small numbers, it seems about one out of three numbers $n=pq,$ twelve out of the smallest forty four, with primes $p \equiv q \equiv 1 \pmod 4$ and $(p|q)= (q|p) = 1,$ do give integer solutions to $x^2 – n y^2 = -1.$ The first few are
$$145 = 5 \cdot 29, \; \; x=12, y=1$$
$$445 = 5 \cdot 89, \; \; x=4662, y=221$$
$$901 = 17 \cdot 53, \; \; x=30, y=1$$
$$1145 = 5 \cdot 229, \; \; x=1252, y=37$$
$$1313 = 13 \cdot 101, \; \; x=616, y=17$$
$$1745 = 5 \cdot 349, \; \; x=4052, y=97$$
$$2249 = 13 \cdot 173, \; \; x=197140, y=4157$$
It seems to hold steady at about one out of three, there were $5820$ successes out of the first $18000$ such numbers. There were $99284$ out of the first $300000.$

HOWEVER:

$205 = 5 \cdot 41,$ and there are no integer solutions to
$$x^2 – 205 y^2 = -1.$$ More to the point, there are no integer solutions to
$$x^2 + x y – 51 y^2 = -1.$$
With reference to the second screen capture below, with $x=20$ and $y=3,$
$$20^2 + 20 \cdot 3 – 51 \cdot 3^2 = 1.$$

Or to
$$x^2 + 13 x y – 9 y^2 = -1.$$

$221 = 13 \cdot 17,$ and there are no integer solutions to
$$x^2 – 221 y^2 = -1.$$ There are also no integer solutions to
$$x^2 + x y – 55 y^2 = -1.$$
With reference to the third screen capture below, with $x=7$ and $y=1,$
$$7^2 + 7 \cdot 1 – 55 \cdot 1^2 = 1.$$

Or to
$$x^2 + 13 x y – 13 y^2 = -1.$$

Screen captures from http://www.numbertheory.org/php/unit.html

Ummm. here is the 205 thing in my language:

=========================================
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$./indefCycle 1 1 -51 0 form 1 1 -51 delta 0 1 form -51 -1 1 delta 6 2 form 1 13 -9 -1 -6 0 -1 To Return -1 6 0 -1 0 form 1 13 -9 delta -1 ambiguous 1 form -9 5 5 delta 1 2 form 5 5 -9 delta -1 ambiguous 3 form -9 13 1 delta 13 4 form 1 13 -9 form 1 x^2 + 13 x y -9 y^2 minimum was 1rep x = 1 y = 0 disc 205 dSqrt 14.317821063 M_Ratio 205 Automorph, written on right of Gram matrix: 2 27 3 41 =========================================  And here is 221 ========================================= jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$./indefCycle 1 1 -55 0 form 1 1 -55 delta 0 1 form -55 -1 1 delta 6 2 form 1 13 -13 -1 -6 0 -1 To Return -1 6 0 -1 0 form 1 13 -13 delta -1 ambiguous 1 form -13 13 1 delta 13 ambiguous 2 form 1 13 -13 form 1 x^2 + 13 x y -13 y^2 minimum was 1rep x = 1 y = 0 disc 221 dSqrt 14.866068747 M_Ratio 221 Automorph, written on right of Gram matrix: -1 -13 -1 -14 =========================================  My Theorem 1 is Theorem 11.5.5 on page 279 of Alaca and Williams. My Theorem 2 is Theorem 11.5.7 on page 286 of Alaca and Williams. It turns out Theorem 2 is due to Dirichlet, 1834. I keep forgetting why$x^2 + xy – k y^2$dominates$u^2 – (4k+1)v^2.$One step: take$x = u-v, y = 2v.$Then $$x^2 + xy – k y^2 = u^2 – 2 uv + v^2 + 2 u v – 2 v^2 – 4 k v^2 = u^2 – (4k+1) v^2.$$ THEOREM 1: With prime$p \equiv 1 \pmod 4,$there is always a solution to $$x^2 – p y^2 = -1$$ in integers. The proof is from Mordell, Diophantine Equations, pages 55-56. PROOF: Take the smallest integer pair$T>1,U >0$such that $$T^2 – p U^2 = 1.$$ We know that$T$is odd and$U$is even. So, we have the integer equation $$\left( \frac{T+1}{2} \right) \left( \frac{T-1}{2} \right) = p \left( \frac{U}{2} \right)^2.$$ We have $$\gcd \left( \left( \frac{T+1}{2} \right), \left( \frac{T-1}{2} \right) \right) = 1.$$ Indeed, $$\left( \frac{T+1}{2} \right) – \left( \frac{T-1}{2} \right) = 1.$$ There are now two cases, by unique factorization in integers: $$\mbox{(A):} \; \; \; \left( \frac{T+1}{2} \right) = p a^2, \; \; \left( \frac{T-1}{2} \right) = b^2$$ $$\mbox{(B):} \; \; \; \left( \frac{T+1}{2} \right) = a^2, \; \; \left( \frac{T-1}{2} \right) = p b^2$$ Now, in case (B), we find that$(a,b)$are smaller than$(T,U),$but$T \geq 3, a > 1,$and$a^2 – p b^2 = 1.$This is a contradiction, as our hypothesis is that$(T,U)$is minimal. As a result, case (A) holds, with evident $$p a^2 – b^2 = \left( \frac{T+1}{2} \right) – \left( \frac{T-1}{2} \right) = 1,$$ so $$b^2 – p a^2 = -1.$$ THEOREM 2: With primes$p \neq q,$with$p \equiv q \equiv 1 \pmod 4$and Legendre$(p|q)=(q|p) = -1,$there is always a solution to $$x^2 – pq y^2 = -1$$ in integers. The proof is from Mordell, Diophantine Equations, pages 55-56. PROOF: Take the smallest integer pair$T>1,U >0$such that $$T^2 – pq U^2 = 1.$$ We know that$T$is odd and$U$is even. So, we have the integer equation $$\left( \frac{T+1}{2} \right) \left( \frac{T-1}{2} \right) = pq \left( \frac{U}{2} \right)^2.$$ We have $$\gcd \left( \left( \frac{T+1}{2} \right), \left( \frac{T-1}{2} \right) \right) = 1.$$ There are now four cases, by unique factorization in integers: $$\mbox{(1):} \; \; \; \left( \frac{T+1}{2} \right) = a^2, \; \; \left( \frac{T-1}{2} \right) = pq b^2$$ $$\mbox{(2):} \; \; \; \left( \frac{T+1}{2} \right) = p a^2, \; \; \left( \frac{T-1}{2} \right) = q b^2$$ $$\mbox{(3):} \; \; \; \left( \frac{T+1}{2} \right) = q a^2, \; \; \left( \frac{T-1}{2} \right) = p b^2$$ $$\mbox{(4):} \; \; \; \left( \frac{T+1}{2} \right) = pq a^2, \; \; \left( \frac{T-1}{2} \right) = b^2$$ Now, in case (1), we find that$(a,b)$are smaller than$(T,U),$but$T \geq 3, a > 1,$and$a^2 – pq b^2 = 1.$This is a contradiction, as our hypothesis is that$(T,U)$is minimal. In case$(2),$we have $$p a^2 – q b^2 = 1.$$ $$p a^2 \equiv 1 \pmod q,$$ so$a$is nonzero mod$q,$then $$p \equiv \left( \frac{1}{a} \right)^2 \pmod q.$$ This contradicts the hypothesis$(p|q) = -1.$In case$(3),$we have $$q a^2 – p b^2 = 1.$$ $$q a^2 \equiv 1 \pmod p,$$ so$a$is nonzero mod$p,$then $$q \equiv \left( \frac{1}{a} \right)^2 \pmod p.$$ This contradicts the hypothesis$(q|p) = -1.$As a result, case (4) holds, with evident $$pq a^2 – b^2 = \left( \frac{T+1}{2} \right) – \left( \frac{T-1}{2} \right) = 1,$$ so $$b^2 – pq a^2 = -1.$$ The viewpoint of real quadratic fields an norms of fundamental units is more concerned with$x^2 + xy – k y^2,$where$4k+1 = pq$in the second theorem. However, we showed above that the existence of a solution to$u^2 – (4k+1)v^2 = -1$gives an immediate construction for a solution to$x^2 + xy – k y^2=-1,$namely$x=u-v, y=2v.$EXTRA David Speyer reminded me of something Kaplansky wrote out for me, years and years ago. From David’s comments, I now understand what Kap was trying to show me. If$x^2 + x y – 2 k y^2 = -1,$then$(2x+y)^2 – (8k+1)y^2 = -4. $This is impossible$\pmod 8$unless$y$is even, in which case$(x + \frac{y}{2})^2 – (8k+1) \left( \frac{y}{2}\right)^2 = -1.$There is more to it when we have$x^2 + x y – k y^2 = -1$with odd$k.\$ Take
$$u = \frac{ 2 x^3 +3 x^2 y + (6k+3)x y^2 + (3k+1)y^3}{2},$$
$$v = \frac{3 x^2 y + 3 x y^2 + (k+1)y^3}{2}.$$
$$u^2 – (4k+1) v^2 = -1,$$ since
$$u^2 – (4k+1) v^2 = \left( x^2 + x y – k y^2 \right)^3.$$