三角数tn=n(n+1)/2が平方数m^2ならば・・・

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

n(n+1)=2m^2,n^2+n=2m^2,n=2m^2-n^2

tn,Tn+2m,t3n+4m+1は等比数列をなす。

n(n+1)(3n+4m+1)(3n+4m+1)={(n+2m)(n+2m+1)}^2

・・・nを消去できないので、この方法では証明は無理

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

P＝(α^k+β^k)/2

q=(α^k-β^k)/2√2

α=3+2√2，β=3-2√2 ,α＋β=6,αβ=1

n={(α^k+β^k)/4-1/2}

m={(α^k-β^k)/4√2}

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

n(n+1)={(α^k+β^k)/4-1/2}{(α^k+β^k)/4+1/2}

={(α^k+β^k)/4}^2-1/4

={(α^2k+β^2k+2)/16}-1/4

={(α^2k+β^2k-2)/16}=2m^2

2m^2=2{(α^k-β^k)/4√2}^2={(α^k-β^k)/4}^2

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

3n={(3α^k+3β^k)/4-3/2}

4m={(2√2α^k-２√2β^k)/4}

3n+4m=(α^k+1+β^k+1)/4-3/2

t3n+4m+1=(3n+4m+1)(3n+4m+2)/2={(α^k+1+β^k+1)/4-1/2}{(α^k+1+β^k+1)/4+1/2}/2

={(α^k+1+β^k+1)/4}^2-1/4}/2

={(α^2k+2+β^2k+2+2)/32-1/8

={(α^2k+2+β^2k-2)/32

={(α^k+1-β^k-1)^2/32

={(γ^2k+2-δ^2k-2)^2/32

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

n={(α^k+β^k)/4-1/2}

2m={(√2α^k-√2β^k)/4}

γ^2=(1+√2)^2=α、δ^2=(1-√2)^2=β ,γ+δ=2,γδ=-1,γ^2+δ^2=6

n+2m={(1+√2)α^k+(1-√2)β^k)}/4-1/2

n+2m={γ^2k+1+δ^2k+1)}/4-1/2

tn+2m={(γ^2k+1+δ^2k+1)/4}-1/2}{(γ^2k+1+δ^2k+1)/4}-1/2}/2

={(γ^4k+2+δ^4k+2-2)/32-1/8}

={(γ^2k+1-δ^2k-1)^2/32-1/8}

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

tn={(α^k+β^k)/4-1/2}{(α^k+β^k)/4-1/2}/2

={(α^k+β^k)^2/32-1/8}

={(α^2k+β^2k+2)/32-1/8}

={(α^2k-β^2k-2)/32}

={(α^k-β^k)/32}

={(γ^2k-δ^2k)^2/32}

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝

tn・t3n+4m+1＝{(γ^2k-δ^2k)^2/32}{(γ^2k+2-δ^2k-2)^2/32}

={(γ^4k+2-γ^2kδ^2k-2-γ^2k+2δ^2k+δ^4k+2)^2/32^2}

={(γ^4k+2-γ^2-δ^2+δ^4k+2)^2/32^2}

={(γ^4k+2+δ^4k+2-6)/32}^2・・・惜しい

＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝