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So I came up with $b= a+1$ $\Rightarrow$ $ab=a(a+1) = a^2 + a$
So that:
$a^2+b^2 -1$ = $a^2 + (a+1)^2 -1$ = $2a^2 + 2a$ = $2(a^2 + a)$ $\Rightarrow$

$(a,b) = (a,a+1)$ are solutions.

My motivation is for this follow up question:

(b) With $a$ and $b$ as above, what are the possible values of: $$ \frac{a^2 +b^2 −1}{ab} $$

Update

With Will Jagy's computations, it seems that now I must show that the ratio can be any natural number $m\ge 2$, by the proof technique of vieta jumping.

Update

Via Coffeemath's answer, the proof is rather elementary and does not require such technique.

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Do you know the Vieta root jumping technique? –  Calvin Lin Jan 28 '13 at 5:31
    
@CalvinLin No what is it? Thanks –  Rustyn Jan 28 '13 at 5:32
    
Vieta jumping –  Calvin Lin Jan 28 '13 at 5:36
1  
Its a pretty well-known exercise in Olympiad math to prove 2,3 are the only possible values. Simply apply Vieta Jumping as Calvin lin suggests. –  dinoboy Jan 28 '13 at 5:43
    
@CalvinLin Thanks these ideas are very interesting –  Rustyn Jan 28 '13 at 5:43

4 Answers 4

up vote 4 down vote accepted

As mentioned, use Vieta's root jumping technique. We know that $(1, r)$ is a solution to the problem with ratio $ \frac {1^2 + r^2 - 1}{1 \times r} = r$, but has the issue that $a = 1$ violates the condition. We can construct another solution by using the Vieta's root jumping technique.

We wish to solve $\frac {a^2 + X^2 -1 }{a \times X} = R$, which is the quadratic equation $X^2 - raX + a^2 -1 $. If this has a root $b$, then by Vieta's formula, the other root is $ra - b$ (since they sum to $r$). Note that this is independent of the condition that $a < b$.

As such, if we have any solution $(a, b)$ to the problem with ratio $R$, then we know that $( ra-b , a)$ is also going to be a solution to the problem with ratio $R$. It remains to check that $0 < ra-b < a$.

Hence, given any solution $(a,b)$, we can, through a finite series of steps, reduce it to $(1, r)$ for some $r$, while keeping the ratio constant. This ratio is then clearly $r$.

share|improve this answer
    
I am having a lot of trouble showing $0<ra-b<a$, any hint? It's probably that I'm just too tired. I feel like some assumptions have to be made about a and b in order to prove that...I'm not too sure. I can't follow the example on wikipedia, because I don't understand how they derived $x_2 < 0 \Rightarrow -kBx_2 \ge k$ (at that step). Initially I thought they got to that step via: $$ -kBx_2 = -(kB)(kB-A)=-(k^2)(B^2-AB) \ge k $$ because $A> B \Rightarrow B^2-AB < 0$ but the problem with that reasoning is I was assuming $A>B>0$, I'm lost –  Rustyn Jan 28 '13 at 8:16
    
Use the fact that $a^2 + b^2 - 1 = Rab$, and $a < b$, so $0 < Rab - b^2 = a^2 - 1 < ab$. This is motivated by looking at the product of the roots (and sometimes we look at the sum of the roots). –  Calvin Lin Jan 28 '13 at 20:53

If you want the ratio $(a^2+b^2-1)/(ab)$ to be equal to $a$, choose $b=a^2-1$.

Then the numerator is $$a^2+b^2-1=a^2+(a^2-1)^2-1=a^4-a^2=a^2(a^2-1),$$ which divided by the denominator $ab=a(a^2-1)$ gives the result $a$.

This answers in the affirmative the question of the "update" of the OP, since here $a$ can be any $m \ge 2$.

I noticed after posting this that Will Jagy mentions the choice $a=r,b=r^2-1$ in a comment. I'll delete this if asked to...

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Very nice,thank you, +1 –  Rustyn Jan 28 '13 at 14:35
    
Just noticed Will Jagy mentions this fact in his comment above re. $a=r,b=r^2-1$. Will include attribution in above answer. –  coffeemath Jan 28 '13 at 14:40

$(3,8)$ is a possible solution. This gives us 24 divides 72, and a value of 3 for (b).

Have you considered that if $ab$ divides $a^2+b^2-1$, then we $ab$ divides $a^2 + b^2 -1 + 2ab$?

This gives us $ab$ divides $(a+b+1)(a+b-1)$.

Subsequently, the question might become easier to work with.

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thanks, +1, I will consider this. –  Rustyn Jan 28 '13 at 5:40
    
This method might move away from the more direct method of using Vieta jumping, as suggested by @calvin-lin. Proceed with caution. –  Poseidonium Jan 28 '13 at 5:42

below, the "ratio" is $$ \frac{a^2 + b^2 - 1}{ab} $$

As you can see, for each fixed $r \geq 3,$ we get an infinite sequence with the same "ratio" $r.$ The first two are $$ r \, ; \; a_0 = 1, \; b_0 = r $$ $$ r \, ; \; a_1 = r, \; b_1 = r^2 - 1 $$ $$ r \, ; \; a_2 = r^2 - 1, \; b_2 = r^3 - 2 r, $$ and generally $$ a_{n+2} = r \cdot a_{n + 1} - a_n \, ; \; b_{n+2} = r \cdot b_{n + 1} - b_n \, ; \; a_{n+1} = b_n. $$

=-=-=-=-=-=-=-=-=-=-=-=-=-=

ratio  a       b
3     144     377
3      21      55
3    2584    6765
3     377     987
3       3       8
3      55     144
3    6765   17711
3       8      21
3     987    2584
4   10864   40545
4      15      56
4     209     780
4    2911   10864
4       4      15
4      56     209
4     780    2911
5     115     551
5      24     115
5    2640   12649
5       5      24
5     551    2640
6    1189    6930
6     204    1189
6      35     204
6       6      35
6    6930   40391
7    2255   15456
7     329    2255
7      48     329
7       7      48
8    3905   30744
8     496    3905
8      63     496
8       8      63
9     711    6319
9      80     711
9       9      80
10      10      99
10     980    9701
10      99     980
11      11     120
11     120    1309
11    1309   14279
12      12     143
12     143    1704
12    1704   20305
13      13     168
13     168    2171
13    2171   28055
14      14     195
14     195    2716
14    2716   37829
15      15     224
15     224    3345
16      16     255
16     255    4064
17      17     288
17     288    4879
18      18     323
18     323    5796
19      19     360
19     360    6821
20      20     399
20     399    7960
21      21     440
21     440    9219
22      22     483
22     483   10604
23      23     528
23     528   12121
24      24     575
24     575   13776
25      25     624
25     624   15575
26      26     675
26     675   17524
27      27     728
27     728   19629
28      28     783
28     783   21896
29      29     840
29     840   24331
30      30     899
30     899   26940
31      31     960
31     960   29729
32    1023   32704
32      32    1023
33    1088   35871
33      33    1088
34    1155   39236
34      34    1155
35    1224   42805
35      35    1224
36      36    1295
37      37    1368
38      38    1443
39      39    1520
40      40    1599
41      41    1680
42      42    1763
43      43    1848
44      44    1935
45      45    2024
46      46    2115
47      47    2208
48      48    2303
49      49    2400
50      50    2499
51      51    2600
52      52    2703
53      53    2808
54      54    2915
55      55    3024
56      56    3135
57      57    3248
58      58    3363
59      59    3480
60      60    3599
61      61    3720
62      62    3843
63      63    3968
64      64    4095
65      65    4224
66      66    4355
67      67    4488
68      68    4623
69      69    4760
70      70    4899
71      71    5040
72      72    5183
73      73    5328
74      74    5475
75      75    5624
76      76    5775
77      77    5928
78      78    6083
79      79    6240
80      80    6399
81      81    6560
82      82    6723
83      83    6888
84      84    7055
85      85    7224
86      86    7395
87      87    7568
88      88    7743
89      89    7920
90      90    8099
91      91    8280
92      92    8463
93      93    8648
94      94    8835
95      95    9024
96      96    9215
97      97    9408
98      98    9603
99      99    9800
100     100    9999
101     101   10200
102     102   10403
103     103   10608
104     104   10815
105     105   11024
106     106   11235
107     107   11448
108     108   11663
109     109   11880
110     110   12099
111     111   12320
112     112   12543
113     113   12768
114     114   12995
115     115   13224
116     116   13455
117     117   13688
118     118   13923
119     119   14160
120     120   14399
121     121   14640
122     122   14883
123     123   15128
124     124   15375
125     125   15624
126     126   15875
127     127   16128
128     128   16383
129     129   16640
130     130   16899
131     131   17160
132     132   17423
133     133   17688
134     134   17955
135     135   18224
136     136   18495
137     137   18768
138     138   19043
139     139   19320
140     140   19599
141     141   19880
142     142   20163
143     143   20448
144     144   20735
145     145   21024
146     146   21315
147     147   21608
148     148   21903
149     149   22200
150     150   22499
151     151   22800
152     152   23103
153     153   23408
154     154   23715
155     155   24024
156     156   24335
157     157   24648
158     158   24963
159     159   25280
160     160   25599
161     161   25920
162     162   26243
163     163   26568
164     164   26895
165     165   27224
166     166   27555
167     167   27888
168     168   28223
169     169   28560
170     170   28899
171     171   29240
172     172   29583
173     173   29928
174     174   30275
175     175   30624
176     176   30975
177     177   31328
178     178   31683
179     179   32040
180     180   32399
181     181   32760
182     182   33123
183     183   33488
184     184   33855
185     185   34224
186     186   34595
187     187   34968
188     188   35343
189     189   35720
190     190   36099
191     191   36480
192     192   36863
193     193   37248
194     194   37635
195     195   38024
196     196   38415
197     197   38808
198     198   39203
199     199   39600
200     200   39999
201     201   40400
202     202   40803
203     203   41208
204     204   41615
205     205   42024
206     206   42435
207     207   42848
208     208   43263
209     209   43680
210     210   44099
211     211   44520
212     212   44943
213     213   45368
214     214   45795
215     215   46224

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

share|improve this answer
    
Wow... interesting evidence that the ratio can be all natural numbers $r \ge 2$ –  Rustyn Jan 28 '13 at 6:17
    
@RustynYazdanpour, yes, I put in some more detail. For each $r,$ put the rows in order (the computer is not very good at that unless you program it yourself), you get certain infinite families: first $(a=1, b=r),$ next $(a=r, b = r^2 -1),$ –  Will Jagy Jan 28 '13 at 6:22
    
@WillJagy Your recurrence can be simplified to $a_{n+1} = b_n$, $b_{n+1} = r b_n - a_n$. This arises from Vieta's root jumping –  Calvin Lin Jan 28 '13 at 6:25

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