Pell’s equation and side and diagonal numbers

Jordan Bell
October 30, 2016

1 Side and diagonal numbers

Heath [17, pp. 117–118]:

a1=1,d1=1.
an=an-1+dn-1,dn=2an+1+dn+1.
dn2-2an2 =4an-12+4an-1dn-1+dn-12-2(an-12+2an-1dn-1+dn-12)
=4an+12+4an+1dn+1+dn+12-2an-12-4an-1dn-1-2dn-12
=2an-12-dn-12
=-(dn-12-2an-12).

As d12-2a12=-1,

dn2-2an2=(-1)n.
(andn)=(1121)(an-1dn-1)=A(an-1dn-1).
(an+1dn+1)=An(11).

A=PDP-1.

D=(1-2001+2),P=(-121211),P-1=(-12121212).
An=((1-2)n2+(1+2)n2-(1-2)n22+(1+2)n22-(1-2)n2+(1+2)n2(1-2)n2+(1+2)n2.).
(an+1dn+1)=(12(1-2)n-(1-2)n22+12(1+2)n+(1+2)n2212(1-2)n-(1-2)n2+12(1+2)n+(1+2)n2).

Thus

(a2d2)=(23),(a3d3)=(57),(a4d4)=(1217),(a5d5)=(2941).

2 Diophantus

If x2-Ay2=1 and y=m(x+1) for rational m, then y2=m2(x2+2x+1), then x2-Am2(x2+2x+1)=1, then (Am2-1)x2+2Am2x+Am2+1=0. Write (x+1)(px+q)=(Am2-1)x2+2Am2x+Am2+1. Then px2+(p+q)x+q=(Am2-1)x2+2Am2x+Am2+1. Then p=Am2-1, q=Am2+1, and so p+q=Am2-1+Am2+1=2Am2. Thus if x-1 then px+q=0 and hence (Am2-1)x+Am2+1=0. Hence (Am2-1)x=-(Am2+1) and so x=-Am2+1Am2-1. Thus

y=m(x+1)=m(-Am2+1Am2-1+1)mAm2-1(-Am2-1+Am2-1)=-2mAm2-1.

Therefore for rational m,

x=-Am2+1Am2-1,y=-2mAm2-1

satisfy x2-Ay2=1. cf. Heath [17, pp. 68–69], Nesselmann [24, p. 331].

Diophantus V.11: 30x2+1=y2. Say y=5x+1. y2=25x2+10x+1. Then 5x2-10x=0, so x=0 or 5x-10=0, i.e. x=0 or x=2. Hence x=0,y=1 and x=2,y=11 satisfy 30x2+1=y2.

Diophantus V.14 [17, pp. 211–212]. 34y2+1=x2. Say x=6y-1. x2=36y2-12y+1. Then 2y2-12y=0, i.e. y(y-6)=0 so y=0 or y=6. Then x=-1,y=0 and x=35,y=6 satisfy 34y2+1=x2.

3 Fermat

Fermat, February 1657 [31, p. 29]:

Given any number not a square, then there are an infinite number of squares which, when multiplied by the given number, make a square when unity is added.
Example. Given 3, a nonsquare number; this number multiplied by the square number 1, and 1 being added, produces 4, which is a square.
Moreover, the same 3 multiplied by the square 16, with 1 added makes 49, which is a square.
And instead of 1 and 16, an infinite number of squares may be found showing the same property; I demand, however, a general rule, any number being given which is not a square.
It is sought, for example, to find a square which when multiplied into 149, 109, 433, etc., becomes a square when unity is added.

4 Wallis

Wallis [1, p. 546].

Stedall [29]

5 Brouncker

Weil [32, pp. 92–99].

6 Ozanam

Ozanam [25, pp. 503–516], Liv. III, Quest. XXVI.

7 Continued fractions

Let

[a0,a1]=a0+1a1

and

[a0,,an-1,an]=[a0,,an-2,an-1+1an].

Then for 1m<n,

[a0,,an]=[a0,,am-1,[am,,an]].

Define

p0=a0,q0=1,p1=a1a0+1,q1=a1

and for n2,

pn=anpn-1+pn-2,qn=anqn-1+qn-2.

Hardy and Wright [13, p. 130, Theorem 149]. For n0,

[a0,,an]=pnqn.

For n1,

pnqn-1-pn-1qn=(-1)n-1.

For n2,

pnqn-2-pn-2qn=(-1)nan.

For n0 let

an=[an,an+1,].

For x=[a0,a1,],

x=anpn-1+pn-2anqn-1+qn-2,n2.

Hardy and Wright [13, p. 144, Theorem 176]. A continued fraction [a0,a1,] is said to be periodic if there is some L0 and some k1 such that al+k=al for all lL.

Theorem 1.

If x=[a0,a1,] is a periodic continued fraction, then x is a quadratic surd.

Proof.

Let

aL,,aL+k-1¯=[aL,aL+1,]=aL.

Thus

[a0,,aL-1,aL,aL+1,] =[a0,,aL-1,aL]
=[a0,,aL-1,aL,,aL+k-1¯].

As aL+k=aL,aL+k+1=aL+1,,

aL =[aL,aL+1,]
=[aL,aL+1,,aL+k-1,aL,aL+1,]
=[aL,aL+1,,aL+k-1,aL].

Let

pq=[aL,aL+1,,aL+k-1],p′′q′′=[aL,aL+1,,aL+k-2].

For t=[aL,aL+1,,aL+k-1,aL],

t=aLp+p′′aLq+q′′=pt+p′′qt+q′′.

Hence qt2+q′′t=pt+p′′, so qt2+(q′′-p)t-p′′=0. For x=[a0,a1,],

x=aLpL-1+pL-2aLqL-1+qL-2.

Then

aL=pL-2-qL-2xqL-1x-pL-1.

Thus, with t=aL,

q(pL-2-qL-2xqL-1x-pL-1)2+(q′′-p)pL-2-qL-2xqL-1x-pL-1-p′′=0.

Then

q(pL-2-qL-2x)2+(q′′-p)(pL-2-qL-2x)(qL-1x-pL-1)-p′′(qL-1x-pL-1)2=0.

Therefore there are integers a,b,c such that

ax2+bx+c=0.

This means that x is a quadratic surd, as x is irrational. ∎

Example. Say x=[3,2,7,4,5,1,12¯]. L=4,k=3.

pq=[5,1,12]=7713,p′′q′′=[5,1]=61.
pL-1qL-1=p3q3=[3,2,7,4]=21562,pL-2qL-2=p2q2=[3,2,7]=5215.

Then

q(pL-2-qL-2x)2+(q′′-p)(pL-2-qL-2x)(qL-1x-pL-1)-p′′(qL-1x-pL-1)2
= 13(52-15x)2+(1-77)(52-15x)(62x-215)-6(62x-215)2
= 50541x2-350444x+607482.

Hence x=[3,2,7,4,5,1,12¯] satisfies

50541x2-350444x+607482=0.

In fact,

x=175222+152250541.

Hardy and Wright [13, p. 144, Theorem 177].

Theorem 2.

If x is a quadratic surd, then the continued fraction of x is periodic.

Example. Say x2=218. 142=196.

218=14+218-14=14+11218-14.
(218-14)(218+14)=218-196=22,1218-14=218+1422.

We do not need to compute the decimal expansion of 218; we merely have to calculate 218+1422. Using 14<218<15,

1218-14=1+218+1422-1=1+218-822.

Then

218=14+11+218-822=14+11+122218-8
(218-8)(218+8)=218-64=154,1218-8=218+8154.

Then

22218-8=22218+176154.

Using that 14<218<15,

22218-8=3+22218+176154-3=3+22218-286154.

Then

218=14+11+13+22218-286154=14+11+13+115422218-286.
(22218-286)(22218+286)=222218-2862=23716,
122218-286=22218+28623716.
15422218-286=3388218+4404423716.

Using 14<218<15,

15422218-286=3+3388218+4404423716-3=3+3388218-2710423716.

Then

218=14+11+13+13+3388218-2710423716=14+11+13+13+1237163388218-27104.
(3388218-27104)(3388218+27104)=33882218-271042=1767695776,
13388218-27104=3388218+271041767695776.
237163388218-27104=237163388218+271041767695776.

Using 14<218<15,

237163388218-27104 =1+237163388218+271041767695776-1
=1+80349808218-11248973121767695776.

Then

218 =14+11+13+13+11+80349808218-11248973121767695776
=14+11+13+13+11+1176769577680349808218-1124897312.
(80349808218-1124897312)(80349808218+1124897312)=142034016204011008.
176769577680349808218-1124897312=176769577680349808218+1124897312142034016204011008.

Using 14<218<15, the floor of the above quantity is 28. Hence

176769577680349808218-1124897312 =28+142034016204011008218-1988476226856154112142034016204011008
=28+218-14.

Then

218 =14+11+13+13+11+128+218-14

Thus for x=218,

x-14=11+13+13+11+128+x-14.

Thus for t=x-14,

t=11+13+13+11+128+t.

Therefore t=[0,1,3,3,1,28¯]. Hence x=14+t=[14,1,3,3,1,28¯]:

218=[14,1,3,3,1,28¯].

8 Euler

Euler, Algebra [10], Part II, Chapter VII.

9 Lagrange

Konen [20, pp. 75–77].

10 Chakravala

Hankel [12, pp. 200–203]

Strachey [30, pp. 36–53]. Dickson [5, pp. 349–350].

Colebrooke [3, pp. 170–184]

Colebrooke [3, pp. 363–372]

Datta and Singh [4, II, pp. 93–99]

Datta and Singh [4, II, pp. 146–161]

Datta and Singh [4, II, pp. 161–172]

Suppose that pn,qn are relatively prime and

Aqn2+sn=pn2.

If d is a common factor of qn and sn then dpn2, so d is a common factor of pn2 and qn2, which implies that pn and qn have a common factor, a contradiction. Therefore qn and sn are relatively prime. Because qn and sn are relatively prime, by the Kuttaka algorithm there are some ρn,ρn satisfying -qnρn+snρn=pn. For rn=ρn+knsn, rn=ρn+knqn,

-qnrn+snrn =-qn(ρn+knsn)+sn(ρn+knqn)
=-qnρn-knqnsn+snρn+knqnsn
=-qnρn+snρn
=pn.

Take rn<A<rn+|sn|. rn=pn+qnrnsn. Let

qn+1=rn,pn+1=pnqn+1-1qn,sn+1=pn+12-Aqn+12.

Example. 69y2+1=x2. A=69.

Aq02+s0=p02: p0=8,q0=1,s0=-5.

p0+q0ρ0=ρ0s0 is equivalent to 8+ρ0=-5ρ0. It is satisfied by ρ0=-8,ρ0=0. Take r0=-8-5k0=7. r0=p0+q0r0s0=8+17-5=-3.

q1=-3.

p1=p0q1-1q0=8-3-11=-25.
s1=p12-Aq12=4.

p1+q1ρ1=ρ1s1 is equivalent to -25-3ρ1=4ρ1. This is satisfied by ρ1=1,ρ1=-7. Take r1=1+4k1=5. Then r1=p1+q1r1s1=-25-354=-10.

q2=-10.

p2=p1q2-1q1=-25-10-1-3=-83.
s2=p22-Aq22=-11.

p2+q2ρ2=ρ2s2 is equivalent to -83-10ρ2=-11ρ2. This is satisfied by ρ2=6,ρ2=13. Take r2=6-11k2=6. Then r2=p2+q2r2s2=-83-106-11=13.

q3=13.

p3=p2q3-1q2=-8313-1-10=108.
s3=p32-Aq32=3.

p3+q3ρ3=ρ3s3 is equivalent to 108+13ρ3=3ρ3. This is satisfied by ρ3=0,ρ3=36. Take r3=36+3k3=6. Then r3=p3+q3r3s3=108+1363=62.

q4=62.

p4=p3q4-1q3=10862-113=515.
s4=p42-Aq42=-11.

p4+q4ρ4=ρ4s4 is equivalent to 515+62ρ4=-11ρ4. This is satisfied by ρ4=5,ρ4=-75. Take r4=5-11k4=5. Then r4=p4+q4r4s4=515+625-11=-75.

q5=-75.

p5=p4q5-1q4=515-75-162=-623.
s5=p52-Aq52=4.

p5+q5ρ5=ρ5s5 is equivalent to -623-75ρ5=4ρ5. This is satisfied by ρ5=3,ρ5=-212. Take r5=3+4k5=7. Then r5=p5+q5r5s5=-623-7574=-287.

q6=-287.

p6=p5q6-1q5=-623-287-1-75=-2384.
s6=p62-Aq62=-5.

p6+q6ρ6=ρ6s6 is equivalent to -2384-287ρ6=-5ρ6. This is satisfied by ρ6=3,ρ6=649. Take r6=3-5k=8. Then r6=p6+q6r6s6=-2384-2878-5=936.

q7=936.

p7=p6q7-1q6=-2384936-1-287=7775.
s7=p72-Aq72=1.

Therefore

77752-699362=1.

Thus 697775936.

Example. 91y2+1=x2. A=91.

Aq02+s0=p02: p0=10, q0=1, s0=9.

p0+q0ρ0=ρ0s0 is equivalent to 10+ρ0=9ρ0. This is satisfied by ρ0=-10, ρ0=0. Take r0=-10+9k0=8. Then r0=p0+q0r0s0=10+189=2.

q1=2.

p1=p0q1-1q0=102-11=19.
s1=p12-Aq12=-3.

p1+q1ρ1=ρ1s1 is equivalent with 19+2ρ1=-3ρ1. This is satisfied by ρ1=1,ρ1=-7. Take r1=1-3k1=7. Then r1=p1+q1r1s1=19+27-3=-11.

q2=-11.

p2=p1q2-1q1=19-11-12=-105.
s2=p22-Aq22=14.

p2+q2ρ2=ρ2s2 is equivalent with -105-11ρ2=14ρ2. This is satisfied by ρ2=7,ρ2=-13. Take r2=7,r2=-13.

q3=-13.

p3=p2q3-1q2=-105-13-1-11=-124.
s3=p32-Aq32=-3.

p3+q3ρ3=ρ3s3 is equivalent with -124-13ρ3=-3ρ3. This is satisfied by ρ3=2,ρ3=50. Take r3=2-3k3=8. Then r3=p3+q3r3s3=-124-138-3=76.

q4=76.

p4=p3q4-1q3=-12476-1-13=725.
s4=p42-Aq42=9.

p4+q4ρ4=ρ4s4 is equivalent with 725+76ρ4=9ρ4. This is satisfied by ρ4=1,ρ4=89. Take r4=1, r4=89.

q5=89.

p5=p4q5-1q4=72589-176=849.
s5=p52-Aq52=-10.

p5+q5ρ5=ρ5s5 is equivalent with 849+89ρ5=-10ρ5. This is satisfied by ρ5=9,ρ5=-165. Take r5=9,r5=-165.

q6=-165.

p6=p5q6-1q5=849-165-189=-1574.
s6=p62-Aq62=1.

Therefore

15742-911652=1.

Thus 911574165.

Example. 109y2+1=x2. A=109.

Ay02+s0=x02: x0=10, y0=1, s0=-9.

x0+y0ρ0=s0ρ0 is equivalent to 10+ρ0=-9ρ0. This is satisfied by ρ0=-10, ρ0=0. Take r0=-10+9k0=8. Then r0=x0+y0r0s0=10+18-9=-2.

y1=-2.

x1=x0y1-1y0=10-2-11=-21.
s1=x12-Ay12=5.

x1+y1ρ1=ρ1s1 is equivalent with -21-2ρ1=5ρ1. This is satisfied by ρ1=2,ρ1=-5. Take r1=2+5k1=7. Then r1=x1+y1r1s1=-21-275=-7.

y2=-7.

x2=x1y2-1y1=-21-7-1-2=-73.
s2=x22-Ay22=-12.

x2+y2ρ2=ρ2s2 is equivalent with -73-7ρ2=-12ρ1. This is satisfied by ρ2=5,ρ2=9. Take r2=5,r2=9.

y3=9.

x3=x2y3-1y2=-739-1-7=94.
s3=x32-Ay32=7.

x3+y3ρ3=ρ3s3 is equivalent with 94+9ρ3=7ρ3. This is satisfied by ρ3=2,ρ3=16. Take r3=2+7k3=9. Then r3=x3+y3r3s3=94+997=25.

y4=25.

x4=x3y4-1y3=9425-19=261.
s4=x42-Ay42=-4.

x4+y4ρ4=ρ4s4 is equivalent with 261+25ρ4=-4ρ4. This is satisfied by ρ4=3,ρ4=-84. Take r4=3-4k4=7. Then r4=x4+y4r4s4=261+257-4=-109.

y5=-109.

x5=x4y5-1y4=261-109-125=-1138.
s5=x52-Ay52=15.

x5+y5ρ5=ρ5s5 is equivalent with -1138-109ρ5=15ρ5. This is satisfied by ρ5=8,ρ5=-134. Take r5=8,r5=-134.

y6=-134.

x6=x5y6-1y5=-1138-134-1-109=-1399.
s6=x62-Ay62=-3.

x6+y6ρ6=ρ6s6 is equivalent with -1399-134ρ6=-3ρ6. This is satisfied by ρ6=1,ρ6=511. Take r6=1-3k6=10. Then r6=x6+y6r6s6=-1399-13410-3=913.

y7=913.

x7=x6y7-1y6=-1399913-1-134=9532.
s7=x72-Ay72=3.

x7+y7ρ7=ρ7s7 is equivalent with 9532+913ρ7=3ρ7. This is satisfied by ρ7=2,ρ7=3786. Take r7=2+3k7=8. Then r7=x7+y7r7s7=9532+91383=5612.

y8=5612.

x8=x7y8-1y7=95325612-1913=58591.
s8=x82-Ay82=-15.

x8+y8ρ8=ρ8s8 is equivalent with 58591+5612ρ8=-15ρ8. This is satisfied by ρ8=7,ρ7=-6525. Take r8=7,r8=-6525.

y9=-6525.

x9=x8y9-1y8=58591-6525-15612=-68123.
s9=x92-Ay92=4.

x9+y9ρ9=ρ9s9 is equivalent with -68123-6525ρ9=4ρ9. This is satisfied by ρ9=1,ρ9=-18662. Take r9=1+4k9=9. Then r9=x9+y9r9s9=-68123-652594=-31712.

y10=-31712.

x10=x9y10-1y9=-68123(-31712)-1-6525=-331083.
s10=x102-Ay102=-7.

x10+y10ρ10=ρ10s10 is equivalent with -331083-31712ρ10=-7ρ10. This is satisfied by ρ10=5,ρ10=69949. Take r10=5,r10=69949.

y11=69949.

x11=x10y11-1y10=-33108369949-1-31712=730289.
s11=x112-Ay112=12.

x11+y11ρ11=ρ11s11 is equivalent with 730289+69949ρ11=12ρ11. This is satisfied by ρ11=7,ρ11=101661. Take r11=5,r10=101661.

y12=101661.

x12=x11y12-1y11=730289101661-169949=1061372.
s12=x122-Ay122=-5.

x12+y12ρ12=ρ12s12 is equivalent with 1061372+101661ρ12=-5ρ12. This is satisfied by ρ12=3,ρ12=-273271. Take r12=3-5k12=8. Then r12=x12+y12r12s12=1061372+1016618-5=-374932.

y13=-374932.

x13=x12y13-1y12=1061372(-374932)-1101661=-3914405.
s13=x132-Ay132=9.

x13+y13ρ13=ρ13s13 is equivalent with -3914405-374932ρ13=9ρ13. This is satisfied by ρ13=1,ρ13=-476593. Take r13=1+9k13=10. Then r13=x13+y13r13s13=-3914405-374932109=-851525.

y14=-851525.

x14=x13y14-1y13=-3914405(-851525)-1-374932=-8890182.
s14=x142-Ay142=-1.

x14+y14ρ14=ρ14s14 is equivalent with -8890182-851525ρ14=-ρ14. This is satisfied by ρ14=0,ρ14=8890182. Take r14=-k14=10. Then r14=x14+y14r14s14=-8890182-85152510-1=17405432.

y15=17405432.

x15=x14y15-1y14=-889018217405432-1-851525=181718045.
s15=x152-Ay152=9.

r15=8,r15=35662389.

y16=35662389.

x16=x15y16-1y15=3566238935662389-117405432=372326272.
s16=x162-Ay162=-5.

ρ16=2,ρ2=-88730210. r16=2-5k16=7. Then r16=-124392599.

y17=-124392599.

x17=-1298696861.
s17=12.

r18=5,r18=-160054988.

y18=-160054988.

x18=-1671023133.
s18=-7.

ρ19=2,ρ19=284447587. Take r19=2-7k19=9. Then r19=444502575.

y19=444502575.

x19=4640743127.
s19=4.

ρ20=3,ρ20=1493562713. Take r20=3+4k20=7. Then r20=1938065288.

y20=1938065288.

x20=20233995641.
s20=-15.

r21=8,r21=-2382567863.

y21=-2382567863.

x21=-24874738768.
s21=3.

ρ22=1,ρ22=-9085768877. Take r22=1+3k22=10. Then r22=-16233472466.

y22=-16233472466.

x22=.-169482428249.
s22=-3.

ρ23=2,ρ23=67316457727. Take r23=2-3k23=8. Then r23=99783402659.

y23=99783402659.

x23=1041769308262.
s23=15.

r24=7,r24=116016875125.

y24=116016875125.

x24=1211251736511.
s24=-4.

ρ25=1,ρ25=-331817152909. Take r25=1-4k25=9. Then r25=-563850903159.

y25=-563850903159.

x25=.-5886776254306.
s25=7.

r26=5,r26=-1243718681443.

y26=-1243718681443.

x26=-12984804245123.
s26=.-12.

r27=7,r27=1807569584602.

y27=1807569584602.

x27=18871580499429.
s27=5.

ρ28=3,ρ28=4858857850647. Take r28=3+5k28=8. Then r28=6666427435249.

y28=6666427435249.

x28=69599545743410.
s28=-9.

ρ29=1,ρ29=-8473997019851. Take r29=1+9k29=10. Then r29=-15140424455100.

y29=-15140424455100.

x29=-158070671986249.
s29=1.

Therefore

1580706719862492-109151404244551002=1.

Thus 10915807067198624915140424455100.

References

  • [1] P. Beeley and C. J. Scriba (Eds.) (2005) The Correspondence of John Wallis, Volume II (1660–September 1668). Oxford University Press. Cited by: §4.
  • [2] W. E. Clark (1930) The Āryabhaṭīya of Āryabhaṭa. University of Chicago Press.
  • [3] H. T. Colebrooke (1817) Algebra, with arithmetic and mensuration, from the Sanscrit of Brahmegupta and Bháscara. John Murray, London. Cited by: §10, §10.
  • [4] B. Datta and A. N. Singh (1962) History of Hindu Mathematics: A Source Book. Parts I and II. Asia Publishing House, Bombay. Cited by: §10, §10, §10.
  • [5] L. E. Dickson (2005) History of the theory of numbers, volume II: Diophantine analysis. Dover Publications. Cited by: §10.
  • [6] K. Elfering (1975) Die Mathematik des Aryabhata I. Text, Übersetzung aus dem Sanskrit und Kommentar. Wilhelm Fink Verlag, München.
  • [7] L. Euler (1738) De solutione problematum Diophantaeorum per numeros integros. Commentarii academiae scientiarum imperialis Petropolitanae 6, pp. 175–188. Note: E29, Leonhardi Euleri Opera omnia I.2, pp. 6–17
  • [8] L. Euler (1764) De resolutione formularum quadraticarum indeterminatarum per numeros integros. Novi commentarii academiae scientiarum imperialis Petropolitanae 9, pp. 3–39. Note: E279, Leonhardi Euleri Opera omnia I.2, pp. 576–611
  • [9] L. Euler (1767) De usu novi algorithmi in problemate Pelliano solvendo. Novi commentarii academiae scientiarum imperialis Petropolitanae 11, pp. 28–66. Note: E323, Leonhardi Euleri Opera omnia I.3, pp. 73–111
  • [10] L. Euler (1984) Elements of algebra. Springer. Note: Translated by Rev. John Hewlett Cited by: §8.
  • [11] D. Fowler (1999) The mathematics of Plato’s Academy: a new reconstruction. second edition, Clarendon Press, Oxford.
  • [12] H. Hankel (1874) Zur Geschichte der Mathematik in Alterthum und Mittelalter. B. G. Teubner, Leipzig. Cited by: §10.
  • [13] G. H. Hardy and E. M. Wright (1979) An introduction to the theory of numbers. Fifth edition, Clarendon Press, Oxford. Cited by: §7, §7, §7.
  • [14] T. L. Heath (1956) The thirteen books of Euclid’s Elements, volume I: Books I-II. second edition, Dover Publications.
  • [15] T. L. Heath (1956) The thirteen books of Euclid’s Elements, volume II: Books III-IX. second edition, Dover Publications.
  • [16] T. L. Heath (1956) The thirteen books of Euclid’s Elements, volume III: Books X-XIII. second edition, Dover Publications.
  • [17] T. L. Heath (1964) Diophantus of Alexandria: a study in the history of Greek algebra. second edition, Dover Publications. Cited by: §1, §2, §2.
  • [18] A. Keller (2006) Expounding the Mathematical Seed. Volume 1: The Translation. A Translation of Bhāskara I on the Mathematical Chapter of the Āryabhatīya. Science Networks. Historical Studies, Vol. 30, Birkhāuser.
  • [19] A. Keller (2006) Expounding the Mathematical Seed. Volume 2: The Supplements. A Translation of Bhāskara I on the Mathematical Chapter of the Āryabhatīya. Science Networks. Historical Studies, Vol. 31, Birkhāuser.
  • [20] H. Konen (1901) Geschichte der Gleichung t2-Du2=1. S. Hirzel, Leipzig. Cited by: §9.
  • [21] J. Lagrange (1766–1769) Solution d’un Problême d’Arithmétique. Miscellanea Taurinensia IV, pp. 41–97. Note: Œuvres de Lagrange, tome 1, pp. 671–731
  • [22] J. Lagrange (1767) Sur la solution des Problemes indéterminés du second degré. Mémoires de l’Académie royale des Sciences et Belles-lettres, Berlin XXIII, pp. 165–310. Note: Œuvres de Lagrange, tome 2, pp. 377–535
  • [23] J. Lagrange (1768) Nouvelle Méthode pour résoudre les problemes indéterminés en nombres entiers. Mémoires de l’Académie royale des Sciences et Belles-lettres, Berlin XXIV, pp. 181–250. Note: Œuvres de Lagrange, tome 2, pp. 655–726
  • [24] G. H. F. Nesselmann (1842) Die Algebra der Griechen. G. Reimer, Berlin. Cited by: §2.
  • [25] J. Ozanam (1702) Nouveaux Elemens d’Algebre. George Gallet, Amsterdam. Cited by: §6.
  • [26] R. Rashed (Ed.) (2012) Abū Kāmil. Algèbre et analyse diophantienne: édition, traduction et commentaire. Scientia Graeco-Arabica, Vol. 9, De Gruyter.
  • [27] C. Selenius (1975) Rationale of the chakravāla process of Jayadeva and Bhāskara II. Historia Math. 2 (2), pp. 167–184.
  • [28] K. S. Shukla (1954) Acarya Jayadeva, the mathematician. Gaṇita 5, pp. 1–20.
  • [29] J. A. Stedall (2002) A discourse concerning algebra: English algebra to 1685. Oxford University Press. Cited by: §4.
  • [30] E. Strachey (1813) Bija Ganita: or the Algebra of the Hindus. W. Glendinning, London. Cited by: §10.
  • [31] D. J. Struik (Ed.) (1969) A Source Book in Mathematics, 1200–1800. Harvard University Press. Cited by: §3.
  • [32] A. Weil (1984) Number theory: an approach through history from Hammurapi to Legendre. Birkhäuser. Cited by: §5.