Saurav's duplet

Number Theory Level 3

Find the number of ordered pairs of integers x , y x, y such that

( x y − 13 ) 2 = x 2 + y 2 (xy-13)^2 = x^2 + y^2

Note: Ordered pairs means that ( 1,2 ) and ( 2,1 ) would be considered as different solutions.


The answer is 4.

Saurav Pal by Saurav Pal , 26, India

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1 solution

Saurav Pal
Mar 26, 2014

(xy-13)^2=x^2+y^2...…
(xy-12)^2+25-2xy=x^2+y^2...…
(xy-12)^2=(x+y)^2-25...…
(x+y)^2-(xy-12)^2=25...…
(x+y)^2-(xy-12)^2=13^2-12^2...…OR (x+y)^2-(xy-12)^2=5^2-0^2......
Now we have to make 6 cases...... CASE 1 :- (x+y)=13 and xy-12=12...…
in this case both roots are irrational...….


CASE 2 :- (x+y)=13 and xy-12=-12...…
in this case there will be 2 solutions (13,0) and (0,13)...….

CASE 3 :- (x+y)=-13 and xy=12...…
in this case the roots will be irrational...….

CASE 4:- (x+y)=-13 and xy=-12...…
in this case there will be 2 solutions (-13,0) and (0,-13)...….

CASE 5:- (x+y)=5 and xy-12=0...... in this case the roots will be complex......

CASE 6:- (x+y)=-5 and xy-12=0...... in this case the roots will be complex......

So, there will be total 4 solutions...….

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thoed ache saval dala kar

Shubhendra Singh - 7 years, 2 months ago

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Chup rhe

Saurav Pal - 7 years, 2 months ago

Can you explain why there are only those 4 cases? For example, we also know that 1 3 2 − 1 2 2 = 5 2 − 0 2 13^2 - 12^2 = 5^2 - 0^2 , so why don't we need to consider that too?

Calvin Lin Staff - 7 years, 2 months ago

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Because in the case 5^2-0^2, there won't be any integer solution.

Saurav Pal - 7 years, 2 months ago

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