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rock.freak667

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## Homework Statement

Prove by mathematical induction,or otherwise,that [itex]23^{2n}+31^{2n}+46[/itex] is divisible by 48 for all [itex]n \geq 0[/itex]

## Homework Equations

## The Attempt at a Solution

Assume true for n=N

[tex]23^{2N}+31^{2N}+46=48A[/tex]

[tex] \times (23^2 + 31^2)[/tex]

[tex]23^{2N+2}+31^{2N+2}+46(23^2 + 31^2)+23^2(31^{2N})+31^2(23^{2N})=48A(23^2 + 31^2)[/tex]

[tex]23^{2N+2}+31^{2N+2}+68540+31^{2N}(528+1)+23^{2N}(960+1)=48A(23^2 + 31^2)[/tex]

Simplifying gives me what I need, i.e. things divible by 48, but 68540 is not...any suggestions?