• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Find positive integers a,b,n that satisfy this expression

1. The problem statement, all variables and given/known data
Are there any positive integers n, a and b such that

[tex]96n+88=a^2+b^2[/tex]


2. Relevant equations
3. The attempt at a solution
It resembles the Pythagorean theorem but I'm not sure how that would help me solve it. I factored the LHS

[tex]2^3((2^2)(3)n+11)=a^2+b^2[/tex]

How do I proceed? Thanks.
 

haruspex

Science Advisor
Homework Helper
Insights Author
Gold Member
2018 Award
30,812
4,449
What can you say about perfect squares modulo 4?
 
What can you say about perfect squares modulo 4?
I'm very bad with the whole modulus concept but I can see that a perfect square divided by 4 will always have a remainder of either 0 or 1.
 

haruspex

Science Advisor
Homework Helper
Insights Author
Gold Member
2018 Award
30,812
4,449
0,1 or 2, right?
 
Right. Now which of those could match the LHS (mod 4)?
96 is divisible by 4 (remainder = 0). 88 is also divisible by 4 (remainder = 0). So sum of remainders (0 + 0) = 0.
 

SammyS

Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,095
881
96 is divisible by 4 (remainder = 0). 88 is also divisible by 4 (remainder = 0). So sum of remainders (0 + 0) = 0.
To get the remainder of zero, means that both a and b are even. Correct ?
 
To get the remainder of zero, means that both a and b are even. Correct ?
Yes, they'd both have to be even. Should I substitute, a = 2u, b = 2v ?
 

haruspex

Science Advisor
Homework Helper
Insights Author
Gold Member
2018 Award
30,812
4,449
Yes, they'd both have to be even. Should I substitute, a = 2u, b = 2v ?
Yes, which allows some cancellation. See if you can then repeat the logic.
 
Yes, which allows some cancellation. See if you can then repeat the logic.
96n + 88 = 4u^2 + 4v^2
24n + 22 = u^2 + v^2
(LHS= 2 mod 4 --> both u and v are odd)

24n + 22 = (2w+1)^2 + (2x+1)^2
6n + 5 = w^2 + w + x^2 + x
(LHS = 3 mod 4 --> ??)

If k^2 = 0 mod 4 then either k = 0 mod 4 or k = 2 mod 4
If k^2 = 1 mod 4 then either k = 1 mod 4 or k = 3 mod 4

Is this correct? And please correct me if my notation is wrong. Thanks.
 
w2+w = w(w+1). What does that tell you about w2+w mod 2?
w(w+1)=odd*even=even --> w mod 2 = 0. same goes for x^2 + x. So the RHS is even but the LHS is odd. therefore it's not possible. Is that it?
 

haruspex

Science Advisor
Homework Helper
Insights Author
Gold Member
2018 Award
30,812
4,449
w(w+1)=odd*even=even --> w mod 2 = 0. same goes for x^2 + x. So the RHS is even but the LHS is odd. therefore it's not possible. Is that it?
That's it.
 
Thank you so much haruspex and SammyS for breaking it down and guiding me through the problem. It helped me a lot :)
 

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top