MHB Finding Triples of Positive Integers in Power Equations

  • Thread starter Thread starter anemone
  • Start date Start date
AI Thread Summary
For a given positive integer p, the problem seeks triples (q, x, y) of positive integers where q is relatively prime to p, satisfying the equation (x^2 + y^2)^p = (xy)^q. It is established that if a pth power is also a qth power, it must be a pqth power, leading to the requirement that q must be greater than p. The derived equations indicate that the only solution occurs when q = p + 1 and both x and y equal 2^(p/2), necessitating that p be even. For odd values of p, no solutions exist.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
For a given positive integer $p$, find all triples $(q,\,x,\,y)$ of positive integers, with $q$ relatively prime to $p$, which satisfy $(x^2+y^2)^p=(xy)^q$.
 
Mathematics news on Phys.org
[sp]If a $p$th power is also a $q$th power (with $p$ and $q$ relatively prime), then it must be a $pq$th power. So if $(x^2+y^2)^p=(xy)^q$, then there must be some integer $z$ with $x^2+y^2 = z^q$ and $xy = z^p.$ Solving the second of those equations for $y$ and substituting that into the first equation, you get $$x^2 + \frac{z^{2p}}{x^p} = z^q.$$ Therefore $x^4 - z^qx^2 + z^{2p} = 0,$ a quadratic in $x^2$ with solutions $x^2 = \frac12\bigl(z^q \pm\sqrt{z^{2q} - 4z^{2p}}\bigr).$ For the expression in the square root to be non-negative, we must have $q>p$, so that $x^2 = \frac12\bigl(z^q \pm z^p\sqrt{z^{2(q-p)} - 4}\bigr).$ But the only way for two squares to differ by $4$ is if they are $4$ and $0$. Therefore $z^{2(q-p)} = 4$, from which $q = p+1$ and $z=2$. The solution is then $q= p+1$, $x=y=2^{p/2}$, which implies that $p$ must be even. For odd $p$ there is no solution.[/sp]
 
Last edited:
Opalg said:
[sp]If a $p$th power is also a $q$th power (with $p$ and $q$ relatively prime), then it must be a $pq$th power. So if $(x^2+y^2)^p=(xy)^q$, then there must be some integer $z$ with $x^2+y^2 = z^q$ and $xy = z^p.$ Solving the second of those equations for $y$ and substituting that into the first equation, you get $$x^2 + \frac{z^{2p}}{x^p} = z^q.$$ Therefore $x^4 - z^qx^2 + z^{2p} = 0,$ a quadratic in $x^2$ with solutions $x^2 = \frac12\bigl(z^q \pm\sqrt{z^{2q} - 4z^{2p}}\bigr).$ For the expression in the square root to be non-negative, we must have $q>p$, so that $x^2 = \frac12\bigl(z^q \pm z^{2p}\sqrt{z^{2(q-p)} - 4}\bigr).$ But the only way for two squares to differ by $4$ is if they are $4$ and $0$. Therefore $z^{2(q-p)} = 4$, from which $q = p+1$ and $z=2$. The solution is then $q= p+1$, $x=y=2^{p/2}$, which implies that $p$ must be even. For odd $p$ there is no solution.[/sp]

What a perfect and brilliant solution, Opalg! (Happy)

Well done and thanks for participating! :)
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top