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Math Question

by Edwin
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Nov16-05, 10:17 PM
P: 167
How would you solve the following system of simultaneous equations for t and b?

sin(pi*t) = 0

sin(pi*(t^2 + 35)/(2*t)) = 0

(t^2 + 35)/(2*t) - t/2 - b/2 = 0

t^2/35 +35/t^2 -t/b - b/t = 0

t*b = 35


Edwin G. Schasteen
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Nov16-05, 10:40 PM
Sci Advisor
HW Helper
P: 3,147
For starters, you should recognize the first equation tells you t is an integer. Likewise, the second tells you

[tex]\frac {t^2 + 35}{2t}[/tex]

is also an integer.
Nov16-05, 11:44 PM
P: 167
That is true. But how do you solve for t algebraically? Is it even possible to solve these systems of equations without using a graphing calculator? Is it possible using numerical methods? If so, which methods?



Nov17-05, 12:46 AM
Integral's Avatar
P: 7,315
Math Question

Simple observation and common sense go along ways in this sort of problem. I do not know of any numerical method which will work well. The problem comes when you are restricted to the integers. This is not the natural domain of numerical methods which are planted firmly in the real number line.

As Tide pointed out your first equations restricts you to the integers, the second further restricts you to a small set of integers.

Rewrite your 2nd condition as

[tex] 2n = t + \frac {35} t [/tex]

So the RHS is an even integer, there are only 3 integers which can satisfy this equation. Can you complete the problem?
Nov17-05, 06:45 AM
Sci Advisor
PF Gold
P: 39,344
If you are referring to t, there are 8 integers that will make the right hand side an integer.

If you are referring to n, there are 4 integers that will satisfy it.
Nov17-05, 10:22 AM
Integral's Avatar
P: 7,315
Quote Quote by HallsofIvy
If you are referring to t, there are 8 integers that will make the right hand side an integer.
If you are referring to n, there are 4 integers that will satisfy it.
I can see 4 integers, 1,5,7,35...I missed the 35 before, But 8? Are you counting negitives as well?
Nov17-05, 08:59 PM
P: 167
Thanks for the help guys. Notice that the numbers 5 and 7 are prime factors of 35. What I am actually trying to do find is to find a general method to solve these systems of equations, if possible, for numbers Cp that are composites of two odd prime numbers.

sin(pi*t) = 0

sin(pi*(t^2 + Cp)/(2*t)) = 0

(t^2 + Cp)/(2*t) - t/2 - b/2 = 0

t^2/Cp +Cp/t^2 -t/b - b/t = 0

t*b = Cp

With the domain restriction on t -> D:{1<t<=Cp}

I constructed these systems of equations because the only values of t in the domain of t given above are the odd prime factors of Cp.

This is true no matter how big Cp is. To find a general and efficient method to solve these equations for large Cp would be to find an efficient method to factor large Cp.

I think

[tex] 2n = t + \frac {35} t [/tex]

is one step closer to solving this problem.

By the way the n of interest is always (a+b)/2 where a and b are the odd prime factors of Cp.

For example, (7+5)/2 = 12/2 = 6, 6 is the only interesting even integer n that satisfies the second condition, and with it, you can immediately factor Cp = 35 as follows.

6^2 - 35 = 1

The prime factors of 35 are then just 6 + 1 = 7 and 6 - 1 = 5

The odd prime factors of Cp can always be solved for by the following algorithm.

{[(a+b)/2]^2 - Cp}^(1/2) = m

a = n + m

b = n - m

So the overall drive to solve the systems of equations above for large Cp is to find n which will enable one to factor large values of Cp.

Any further assistance would be most appreciated. I've found so far that these problems equations are difficult to solve, and it is my conjecture that there exists no efficient general method to solve the system of equations above. Please prove me wrong

Best Regards,


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